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2 



A KEY 



TO 



ELEMENTARY STATICS AND 
HYDROSTATICS 



-1 




A KEY 



TO 



ELEMENTARY STATICS AND 

HYDROSTATICS 



BY 



M 



J. HAMBLIN SMITH, M.A. 

OP GONVILLB AND CAIUS COLLRGB, 
AND LATE LECTURER AT ST. PETEK'S COLLEGE, CAMURIDGK 




M » *J V w 



RIVINGTONS 



WATERLOO PLACE, LONDON 

MDCCCLXXXIII 






ELEMENTARY STATICS. 

KEY. 

Examples — I. (p. 25). 

1. Let C be the body. ACB a straight line. 
Then if the three forces act all in the same direction along CBy a 
force of 12 lbs. will be required acting along CA to keep them at rest. 



^ 




c 








B 








3 


4- 


5 




A 




c 








B 




3 






5 


4 




A 


^ 


c 








B 




4 






\ 


7 




A_ 


^ 


c 




— ^— 


^ 


B 



Fio. 1. 

If the forces of 5 lbs. and 4 lbs. act along CB^ and the force of 3 lbs. 
along C-4, a force of 6 lbs. wiU be required acting along CA to keep 
equilibrium ; and so for the other cases. 

2. As in the preceding example, if the forces all act along OB they 
will have a resultant acting along CB of 29 lbs. 

If the forces of II lbs. and 13 lbs. act along CB^ and the force of 5 
lbs. along CA^ thej will have a resultant of 19 lbs. acting along CB ; 
and so for the other cases. 

3. Let X be the measure of the magnitude of the resultant in lbs. 
Then a2= (90)2 + (120)2 

=8100 + 14400=22500; 
.'. x=I50, and the resultant is one of 150 lbs. 

A 



KEY TO ELEMENTARY STATICS. 



4. Here «*= (36)2 + (48)2 

= 1296 + 2304 = 3600; 
.'. «= 60, and the resultant is one of 60 lbs. 

5. As the squares of the numbers 6 and 8, that is, 36 and 64, are 
together equal to 100, which is the square of 10, it is plain from the 
first diagram on page 23 of the Statics, that if the measures of AC, 
AB be 8 and 6, and the measure of AD be 10, the square on AD 
will be equal to the sum of the squares on AC, CD, and, therefore, 
I ACD will be a right angle (Euclid, I. xlyiii) ; whence i BAC is 

also a right angle (Euclid, I. xxix). 

6. Since the squares of 3 and 4, that is, 9 and 16, are together 
equal to 25, which is the square of 5, the forces 3 and 4 will act at 

f^ jQ right angles to each other. Hence if C 

be the point, draw CA, CB containing 
3 and 4 units of length respectively, 
Q complete the parallelogram ACBD, 
Produce DC to E, making CE^DC^b 
units. Then GA, CB, CE will repre- 
£ sent the arrangement of the three 

forces. 




FiO. 2. 



7. Let the measures of the forces be x and ,J3 . x, which numbers 
are in the ratio of 1 : jj3. 

Then a;« + (V3.a;)2= (10)2; 

or, at^ + Zx^^lOO ; or, 4a^=100 ; or, a;=5 ; 
.'. the forces are 5 lbs. and 5 ,J3 lbs. 

8. Construct a diagram as in Example 2 on p. 23, but with 
I BAC-^20\ 

Then 1 2)0^=30% L CDE^^% and CD^WE. 
Hence C^=- CD^ - DE^ 

^CT^ Giy'.^CD^. 
^^ "T" 4 ' 



KEY TO ELEMENTARY STATICS, 



Then if a; b^ the measure of AI>^ 



-v/3.6 



2 
=9 + 25 + 15 V3=34+ 15V3 ; 

.*. «=V34+15V3. 
Or, by Trigonometry, 

x2:«32^.5« + 2 .3.5. co830'=9 + 25 +30 . ^=34 + 16^3. 

9. Construct a diagram as in Example 2 on p. 23. 

Then x2=102 + 7^ + 2 . 10 . -^ 

2 

= 100 + 49 + 70=«219; 
.-. a;«V219. 
Or, by Trigonometry, 

a;«=102 + 7« + 2.10.7.cos60'=100 + 49 + 140.4 = 219. 

2 

10. Let ^-B, AC represent the forces, l BAC being an angle of 
135°. 

Complete the parallelogram ABDC, draw DE perpendicular to ^C 
Then z ^GD= 45* (Eucud, I. xxix.), q q 

and EC=^ DE. 
Also, CD^^DE^ + EC^ 
^2E(P, 

and.-.^C=^. j,^3 

Hence, if aj be the measure of -4D, 
x^=AC^-^CIP-2AC,EC. (Euclid, II. xiii.) 

= 112 + 92-2. 11 . -5^=121 + 81-99 V2. = 202-99 ^2 ; 

.-. x= a/202 -99^2. 
Or, by Trigonometry, 

a;2=ll2 + 92 + 2.11.9.cosl35' 

= 121+81 + 198 X (--j=)=202-99 ^2. 




KEY TO ELEMENTARY STATICS. 



11. Construct a diagram as in Example 2 of p. 23, but having 
L BAC^46% and .*. z DCE^46°. 

CD 
Then, as in Example 10, (^^=—7q' 

Hence, if a; be the measure of J.I> 
x^=AC^ + CI>^ + 2AC,CE 

5 




=4*^ + 52+2.4. 



V2 



= 16 + 25 + 20^2=41+20^2; 

.-. a:= V4I + 20 ^2. 
Or, by Trigonometry, 

x2=42 + 52 + 2.4.5.cos45" 

= 16 + 25 + 40.-^2=41 +20 V2. 

12. This Example is to be worked precisely in the same way as 
Example 2 on p. 23, except that instead of 8 lbs. we must put F to 
represent each of the given forces. Then the magnitude of the 
resultant will be F a/3 in place of 8 ^3 lbs. 

Or, by Trigonometry, 

a;2=jp2 + ^ + 2i^.cos60'' 

= 2F^ + 2F^,^ = ZFK 

13. Let AB^ AG represent the forces acting at A. 
Produce BA to ^, making AE=AB. 
Complete the parallelogram ABDC, AEFC, 

Then, if P and Q be the forces represented by AB and AC, 
and if z BAC^a, and .*. z J^^(7=180°-o, and if B and R' be the 
resultants in each case, 

il2=p2 + Q2 + 2PQ.cosa, 
ir2=.p2+Q2-2PQ.coso; 

.-. jB2+jB'2 = 2(P2+Q2). 

But since DAF is a right-angled triangle, 
FD^^AD^ + AF^; 

.'. (2P)2=i22 + ie^ 

Pio. 6. .-. 2P2=2^, 

which cannot be uoless P=Q, 




KEY TO ELEMENTARY STATICS, 



14. Let D be the point where the vertical line of direction, in 
-which the weight acts, meets a line drawn from A parallel to BF, 

Then the component forces, which are ¥ A 
and the tension of the string BA^ being 
represented by BG and BA^ BD will repre- 
sent the resultant force of 10 lbs. 

Now, DBF is a right angle, and z DCB 
=60°. (Euclid, I. xxix.) 

/.BD^^Z.BC; 

...p-i»,u-!»3^5,u. 




Uiolhs 

Fio. 6. 



Examples— 11. (p. 31). 

1. As in Example 12 of I., we find the resultant of P and P to be 
P . V3 ; and the resultant of Q and Q is mJq^'+Q^^ or Q . a/2~; 

.'. since the resultants are equal, 

P,J3--Q^2, that is, P : Q-^>J2 : V3. 

2. Let P and Q be the forces, and let P, their resultant, be equal 
to P. Draw a diagram as in Art 38. 

Then since AB^AD, .-. z ABD= l ABB. 
Now L ABD=eO% because z 5^40=120° ; 

.-. z^PD and z^DP are each=60°; 

.-. z Pull) ==60° (Euclid, I. xxxn.) 

that is, the triangle ABD is equilateral, and /. P= Q. 

3. (1) The proof depends on the fact that the diagonals of a 
parallelogram bisect each other. See Art 34, 3. 

Hence FB is half the diagonal of a pai^ 
allelogram, of which J.P, DE are the adja- 
cent sides. 

And EF is half the diagonal of a paral- 
lelogram, of which BF, CF are the adjacent 
sides. 

Hence resultant of AE, DE =2PP, 
and resultant of PP, CF=- 2EF ; 

.', since 2PP and 2PP represent equal and opposite forces, forces 
represented by AE, DE, BFy CPare in equilibrium. 




Fio. 7. 



KEY TO ELEMENTARY STATICS, 



(2) The length of a side of the square being 5, the length of the 

diagonal is V/S^ + ^S*, or ^JY. 

Now the resultant of forces B and BJ^ acting at an angle of 45* 
(for the diagonal bisects the angles of the square), is found, as we 
proved in Example 11 of I., thus — 

v" 



4. If ABy iiC represent the two equal forces, and AD the third force, 

Q £ z J?^C=60% because the sum of 

L^BABy CAD=^ 300'. 

Then, as in Example 12 of I., 

AE^^^.AB. 

Hence if ^ be the magnitude of 
each of the equal forces, the magni< 
Pio. 8. tude of the third force is ^3 . F. 




5. (1) Let be the centre of the circle described about the 

triangle ABC, 

A 




Fio. 9. 




Then OA = OB^ OC. 

Now AO passes through 2>, the middle point of BC, by the con- 
ditions of the question. (See Examples — II. 3 (1).) 

Hence (fig. I.) OD is perpendicular to BC (Euclid, III. in.) ; 
.*. AB=AC (Euclid, I. rv.), and therefore ABC is isosceles. 

If 0, the centre of the circle, lies in BC, then OA^OB^OC, and 
the angle BAC^ being the angle in a semicircle, is a right angle ; 
that is, the triangle is ri^t-angled. 



KEY TO ELEMENTARY STATICS. 



7 



(2) The two smaller forces being at right angles, let the measures 
of the three forces, taken in order of magnitude and represented bj 
AB^ ACf AD hex-dfXfX + d, 

Then AE^==AC^+ CE*, 
and/. {x+d)^^a^+{x-'d)^; 

Hencei x^4dy and the measures of the 
forces are Zd, 4d, fid ; that is, the common 
difference d is one-third of the least force. 

N,B. — There is an error in this Example 
in the edition of 1875 : for ffreaUit read 
least but one. 




Pfo. 10. 



6. This is worked in the same way as I. 9, and we get 

«»=4 + 9 + 2.2.4- = 13 + 6-19. 



7. This is worked as L 9, obsenring that the resultant is given, and 
that X must be put for one of the components ; then 

X 



3«=2*+iB« + 2.2. 



2 



or, 9=4 + {B^ + 2x. 
Hence «• + 2a;=5 ; whence «— ,J6 - 1. 



8. Let be the point of application, OE the line pointing to the 
East. O /^ 

Let OAy OB, represent the components, 
so that Zu405=135^ 

Draw AR at right angles to OE. 

Then since i u40iJ= 45% and i ABO B' 
=90", z 0XB=45'. 

.'. if the measure of AO be aJ2.P, the measure of OB=P, and 
since DA = BO, the measure of DA — P. 

Hence DOBA is a parallelogram. 

.'. OD is parallel to A By and is therefore in the direction of the 
North, and OD^AB, and therefore its magnitude is P. 




8 



KEY TO ELEMENTARY STATICS. 



9. Let OAy OB, OC represent the three forces P, 2P, ^3 • P 
acting at 0. 

Complete the parallelogram OADB, 

Then since OD represents 2P, and 

(2P)«=( V3. Ff+F\ I DAO is a right angle ; 

/.since 0D=2,0A, 

zl>0^ = 60'; and.-. z^0i)=30'. 
Hence z^0^=90', 
z BOC^ lbO% 

Pio.12. iA0C^l20\ 




10. Produce RO to any point 2>, and draw D^ parallel to OF, 
meeting OQ in B, 

Then the sides of the triangle DOB being 

Q parallel to the forces P, Q, jR, are proportional 

to them. 

IB ^ Now since z JBOQ=136", .'. zD0^=45^ 

^ Also z 2)^0=90°. 

••. P:Q:B=DB:BO:DO 

= 1 : 1 : V2. 
Fio. 13. 





r^Q 



no. 14. 



il. Construct a diagram as in Example 10. 
Then since P=C=JB, 

i>JB=JB0=02>. 
.-. z2>0^=:60'; 
.-. iROQ = l20\ 
Similarly it may be shown that 
I FOQ= A FOR= 120\ 



KEY TO ELEMENTARY STATICS. 



12. Let ABy ^C be the components, AD the ^ 
resultant perpendicular to AB, 
Now, since z ^^(7=120% .*. z^JBi)=60^ 
Hence in triangle DAB 

ABiBD'.DA^l ;2: V3; 
/. AB.ACiAD^l :2: ^3. 




13. Let ABy.AC represent the components, AC='2ABy and AD 
the resultant. 
Produce BA to E, making AE=AB, 
Then since z 01^=120°, .-. i 2>^J^«60" ; 
and .'., since BE^BD, 

L BDE= L BED=^60°, 
Hence DBE is an equilateral triangle, and DAj 
which bisects the base, is at right angles to BA, 
.-. AD:AB= V3 : 1, or, AD= V3 . P. 




Fia. 16. 



14. Let AB, ^0 be the components, and let AD the resultant 

Produce AB to E, making BE==AB, 
Then AE=AD. 

But, since DB bisects AE at right angles, ° 

AD=^DE; 
.'. ^^D ia an equilateral triangle, A 

and .*. z ^^2) = 60'. Fio. 17. 




15. Let AB, AC he the components, B 
AD the resultant =3 ^C 
Then since iBAC== 136', . '. z DOi = 45^ 
Hence z ^Z>C= z VCA=-46% 
9^d,\ I D AC =90\ 
Then ^5 : ^C= CD\AC=^mJ2: 1. 




10 



KEY TO ELEMENTARY STATICS. 



16. Let AO^ BO be the strings, 
CO the vertical line 

A From any point DiaOC draw DE parallel 
E to OB. 

Then the sides of the triangle DOE are 
parallel to the three forces. 

Now I DOE =30% iDEO^%0\ 
and .-. L EDO^eo\ 
.'. tension of -4 : tension of BO^EO: ED 
= V3:1. 




Fio. 19. 



17. (1) Construct a diagram as in Example 13, and AD the re- 
sultant, being at right angles io AB, makes an angle of 30** 
with AC. 

(2) The components being equal, the resultant bisects the angle 
between them, and this being an angle of 135°, the resultant makes 
an angle of 67^° with each component. 



18. Produce AD to E, making DE^AD. 

p p Join BEf CE, 

Then .'. AD^ED, and BD= CD, 

md I BDE=^ I ADC; 
.'.iBED^iDAC. 
Hence BE is parallel U> AG. 
Again, •.• AD'^ED, and BD^CD, 
A C and i ADB=EDC, 

Fio. 20. /. Z DBA »= Z DCE. 

Hence AB is parallel to EC, 
Thus ABEC is a parallelogram, and AE ia the resultant of AB, 
AC 




KE Y TO ELEMENTAR V ST A TICS. 1 1 



19. Construct a diagram as in Example 10. 

Let jB, P, Q be the order of magnitude, 
B, being the greatest. 

Then ODy DBf BO are in descending order ; 

.-. I OBD, L BOD, L ODBiae in descend- 
ing order. 

.-. z POQ, L QOB, I BOP are in ascend- 
ing order, 

for z OBD + z POQ = 2 right angles, and 

similarly for the others. Fio. 2i. 




20. Take a parallelogram ABCD, having z BAD less than z ABC, 
Now ABD and BAC are two triangles 
having two sides equal, each to each, but 
the included z BAD in the one less than 
the included z ABC in the other, 
.'. BD is less than AC, 
Or, by Trigonometry, 

JJ«=pa + ^ + 2PC.co8a; Fio. 22. 

and as a increases from 0^ to 90% cosa diminishes, 
and as a increases from 90'' to ISO**, cosa increases, but is negative ; 
.', B\a always decreased as a increases. 




21. Let a be the angle between the strings, t the ten- 
sion of each string, w the weight supported. 

Then ti;2= e« + e* + 2«« . coso. 

Now, as the string is lengthened a is decreased, 

*. cosa is increased ; 

and .*., since w remains constant, t is decreased. fio 28. 

The pressure on the nail will not be affected by the length of the 
stripg. 




12 



KEY TO ELEMENTARY STATICS. 



22. Let A be the peg, and let AB^ AC represent the equal forces 
P and P pulling at the ends of the string. 

A Then the strain on the peg is a 

force represented by AD, the 
diagonal of the rhombus ACDB. 

Also, AD-=AB=Aa 
^ ,\ A CD and ABD are equilateral 
D triangles ; 

.-. iCAD^iBAD=eO''; 

.-. iCAB=l20\ 

Fia. 24. 




23. The polygon, the sides of which represent the forces in magni- 
tude and direction, must be a rhombus or a square, for all the sides 
are to be equal Hence, since the opposite sides of the polygon are 
parallel, the forces must be opposite, two and two. 




Fio. 25. 



24. Take F, the middle point of CD, and join 
EF ; then EF is parallel to ^C. 

Now, resultant of AD, AC=2, AF (Example 
18), and resultant of AE, AC^AF; 

'\ resultant of AD, -40= twice resultant of 

AE, AC. 



25. In the first system B is the resultant of P and Q ; in the second 
system jR' is the resultant of P and Q wheit these forces are reversed. 

.*. jR is equal and opposite to B\ 



26. Take the diagram on page 30 of the Statics. 

The force representing the combined eflfect of AB, BC, CD, DE is 
AE, and therefore the forces represented by AB, BC, CD, DE are 
kept in equilibrium by a force represented by a line equal and oppo- 
site to AE. 

Hence X lies in EA produced, so that AX—AE. 




KEY TO ELEMENTARY STATICS. 13 

27. Let be the point loithin the quadrilateral ABCD. 
Now E, F, Gj H, being the points of bisection of the sides of the 
quadrilateral, being joined would form a 
parallelogram, and the diagonals of this 
parallelogram bisect each other in N, 
Then resultant of OA, OD is 2. 0(?, 
and resultant of OB, 0Cia2,0E; 
.-. resultant of OA, OB, OC, OD is 

2 . resultant of OQ, OE, 
But resultant of OC?, JS? is 2 . OiV ; p^^ 2^ 

.-. resultant of OA, OB, OC, 02) is 4. ON, 

The construction and proof are precisely the same when is without 
the quadrilateral. 



Examples— III. (p. 40). 

1. Let AB represent a force of 12 lbs., AC, AD the components. 
Then z CAB^90% L ^150=30^ C B 

and.-. ^-405=60^ 

Then AG. CB:AB=l : 2 : V3 ; 

.*. AC : 12=1 : ^3, or, -4 (7 represents A ^ 

a force of 4 *J3 lbs ; Fio. 27. 

and CB : 12=2 1^3, or, AD represents a force of 8/v/3 lbs. 

2. (1) Let AB represent a force of 10 lbs., AD, AC the compo- 
nents, each making an angle of 30* with AB, and being therefore 
equal. 

Draw BE at right angles to AC 
produced. 

Then i BCE^ i DAC^eO"" ; 
md ,\ BC=2CE. 
Ji^ow A&=^AC^+CB^ + 2AC.CE, 

AC 
or 100^ AC^ + AC^ + 2AC. 

.-. 100=3^ C«, and .-. AC^^-^^AD. 

«5 





14 



KE y TO ELEMENTAR V STA TICS. 




(2) Since i BAD^ l BAQ^m" 



.-. AC^ AD = AB^ 10 Iha. 



Fjo. 29. 



3. Let a be the angle. 

Then, since B^'^F^ + Q^ + 2PQ . cosa 
92=9« + 6» + 2.9.6.coso, 
or 81 =81 +36 + 108 cosa ; 



36 



1 



4. (1) Let AB represent the given force, and let CD, -Si'' repre- 
sent the other forces. 




D 



Pio. 30. 

With centre A, and distance CD, describe a circle. 
With centre B, and distance EF, describe a circle. 
Let M be a point where the circles intersect. 
Join AM, MBy and complete the parallelogram AMBN, 
Then AM, AN represent the component forces. 

(2) Let a and fi be the angles that the 
component forces are to make with the 
given force represented by AB. 

Draw AC making z BAC=a, 
and BD making z DBA =^, 
and let ^ be the intersection of AC, BD. 
Complete the parallelogram AMBN. 
Then AN. AM represent the component 
Fio. 81. forces. 





KEY TO ELEMENTARY STATICS. 15 

5. Complete the parallelogram ACDB. 
is the middle point of AD and BC, 
AC= 13, OC-5, and ^OC is a right angle, 
because BAG is an isosceles triangle. 

Then ^a«=13«- 52-144 ; A 

.-. ^0=12, and /. ^D=24. pio. 82. 



6. First, C lies on the circumference of the 
circle described with A as centre and AB 
as radius. 

Next, D lies on the circumference of the 
circle described with B as centre and BA 
as radius. 




7. If P and Q be the component forces and a the angle between 
their directions, and if Q be reversed, the angle that it makes with F 
is 180** - a. Hence, if B and iSf be the resultants, 

JJ2=P« + ^ + 2PC.cosa, 

;S[2=P2 + ^ + 2P^ . cos (180' - a) 
= pa + C2-2PC.cosa; 
.\B^ + 8^^2(P^ + (y), which is independent of a. 



8. Let T be the tension of each string in the first case, t the tension 
of each string in the second case, W the weight of the picture. 

Then TF»= T* + T« + 2T« . cos60% 

and TF2= ^^ t^ + 2t^ . C08l20' ; 

.-. 2T2 + 2T».J=2e2 + 2«^ x(-i), 

or,3T*=e«; 

.-. T«:^-:l :3; 

.-. r :t =1 : y/3. 



i6 



KEY TO ELEMENTARY STATICS, 



9. Let AQ represent the force of 9 lbs. Then take the formula for 
finding the cosine of one of the angles of a triangle in terms of the 
sides (Trigonometry, Art. 179), 




Fia. 34. 



cos.4 = — ^n^ , and we have 

2oc ' 



C082)/1C= 


144 + 81- 
2xl2x 


-36 
9 " 


225 - 36 
" 216 " 


189 
"216" 


7 
"8 


cos5^D= 


144 + 36- 
2xl2x 


-81 
6 " 


180 - 81 
144 


99 
"144" 


11 
"16 



10. Let Pi and Pa be any two positions of P, and let be the point 
y\ in the arc AGB through which the resultant 
passes when P is in the position Pi. Join 
CP2. 

Then i AP^B^ l AP^, and 
/B I APiC^:' L AP^C, (Euclid, III. xxi.) 

Now since the two forces are the same at 
Pi as at Ps, and act at the same angle, the 
resultant must make with them the same 
angle in both positions. 
.'. PjC must be the direction of the resultant for the position Po. 




11. Let the string lie on the quadrant POQ, and let be the centre 
of the circle, and OC a vertical line. 

Then the forces acting on P are its weight 
P, R the pressure of the semicircle acting 
along the normal OP, and T the tension of 
the string, which acts as a tangent at P. 
Let z COP^a. 
Then T : P=sino : sin90'. (Art. xxxix.) 

.-. T=P.sina. 
SimilarlyT=C . sinGOQ=Q, cosa. 

Q 
.•, P. 8ino=^. cosa, and .•. tana = p- 

FiaS 




KEY TO ELEMENTARY STATICS. 



Hence the eight forces clearly form a 
n equilibrium. 

13. From Art. zltiil 

iF'=(S+4 cos30" + 4 coaeo7, 

+ (6 +■ 4 cob60' + 4 eos30=)' 
-(8 + 2^3 + 2)' + (6 + 2 + 2^/3)' 
-(10 + 2V3)' + (8 + 2V3)* 
= 100 + 40 s/3 + 12 + 64 + 32^3+ 13 
=.188 + 72^3-4(47 + 18^3); 
.■- je=2v'(4T + 18V3). 




Also tand^ 



6 + 2^3 4+ V3 _ (4+ V3)(5- V3 ) 17+ ^3 
'10+2V3"6+V3 (5+ V3)C5- ^3)° 22 



14. Let F be die (^tcd point. 
Take Jtfand iVthe middle points of AB and 
r'D. ThelinejoiningJlfandJrpMBea through 
O, the centre. 

Resultant tAPA, PB=S . FN. 
Resultant of PC, PD~iPM ; 
.: ^waltmtotPA,PB.PC,PD 

=2(re8ultant of Pif, Pitf) 

=2x2.P0. 

= 4P0. 

16. PM represents joint effect of PO, 
OM. 

NQ represents joint effect of OQ, NO ; 
.: PM, NQ cBpreeant joint effect of PQiNM. 
„ „ otAD, AB, 

otAC. 
.: PM, NQ, CA fonn a Bjatem ia- 
equilibrium. 




i8 



KEY TO ELEMENTARY STATICS. 



Examples — IV. (p. 45). 

(1) 20 : 30=^0 : 5, or, 100=3050, /. 5C=3J inches. 

(2) Here P + ^=28, and .'. P : 28-P=3 : 4 ; 

.'. 4P=84-3P, whence P=12, and /. Q=16. 
Then 12 : 16=7-^C : JLC, whence ^(7=4 inches. 

(3) P = JB - C=(14i-i) lbs. = 14 lbs. 

Then 14 : i=PO: 2, and .'. PC=56 inches. 
Hence ^P=58 inches =4 feet 10 inches. 

(4) 3 : 5= JLO- 12 : AC, and .'. ^C=30 inches=2 feet 6 inches. 
(6) 10 : Q=18-6 : 18, and .'. ^=15 lbs. 



M 



5/bs 



>\ 



Fig. 4L 



Examples— V. (p. 52). 



N 



7 /6s 



1. 5:7^CN:CM 

= G2V:6-GZV; 
.-. 30~5CW=7CW; 
.-. 0^=2 J feet, and CAr=3i feet. 



M 



5 m 



TV 



Fio.42. 



N 



7J/6S 



2. CJf :aZV=16:6 
= 3:1. 



M 



C 
IT 



FI0.4S. 



3. P: ^CNiCMy 
aiidP + ^=99; 

.-. P:99-P=7:4, 

.-. 4P=693-7P; 

.-. P=63 lbs., and .*. ^=36 lbs. 



KEY TO ELEMENTARY STATICS. 



^9 



4. C=(16-,12) lbs. -4 lbs. ; 
.-. 12:4-»OiV:lfoot; 
.•.CiV^= 3 feet. 



M 



;2f6s 



C 



Fia. 44. 



5. Ci^=36-CAr; 

.-. 5:7=36-OAf:(7M; 

.-. 5CAf=252-7CM; 

.'. 03fs21 inches. 



M 



C 
IT 



IG. 45. 



6. CM : CN=- 5 : 2, the condition 
that one of the arms is 5 inches 
longer than the other having no 
^ect on their rdaJtive value. 



M 



2 /6s 



M 



Fia. 4«. 



7. Let P be the smaller weight. 

Let X be added to P and taken 
from Q, and let P+x act at Ny and 
Q-xat M, 

Then«Jl^«^. 

Hence — - ^— • 
Hence -^_^^^, 

or,P« + Pa;=^-^ 

or (P+C)a;=^-P', and .-. a:-C--P. 
If we take P the larger weight, x=P- Q. 



vv 



Fia 47. 



N 



J/65 



N 



^ 



20 



KEY TO ELEMENTARY STATICS. 



M 



M 



Fio. 4a. 



Fro. 49. 



N 



8. C7iV=33-a3f; 

.-. 03f :33-C3f=3:8; 

.-. 80M=99-3CM; 

/. (7^=9 inches. 



9. CM:8-CM=3:1; 



/. CAr=24-3CM; 



.-. CM"=6 feet 



M 






vv 



10. Let C be the first position of the fulcronL 

Then03f:12-CM=6:3; 
D G B N .-.30211=72-6031; 

.-. CM'=8 feet. 

Now move the weights of 3 lbs. 

and 6 lbs. to ^ and B, and let D 

A( be the new position of the fulcrum. 

6lbs ThenAB=8feet,andl>5=8--4A 

Pro. 60. and ^D : 8 - ^ D = 6 : 3 ; 

.-. 3^D=48 - e^JD, or u41>=6i feet. 
Now ^0=6 feet, /. 2>C=| feet=8 inches. 



M 



C 
7^ 



N 11. C!af=8in., a2V=38in. 



22 lbs 



.-.P: 22=38:8; 

/. 8P=836 ; 
.-. P-1044 lbs. 



Fio. 51. 



KEY TO ELEMENTARY STATICS. 



21 



12. Let X be the length of ^C in inches. 



ThenCZV:OM=10:15, 



% 



or, ^:x- 6=2:3; a 



3a; 
••. 2-=2x-12, 

or, OSS 24 inches » 2 feet. 



lOlbs 



M 



N 



Fio. 51 



C 
71 



13. C2V:C3f=14:25; 
.-. 0^:36=14:26; 
.-. 26(W=604 ; 



M 



^ ^/4lb8 



A 25l6s 



N 



Fia6& 



7\ 



14. Take N the middle point of AB^ this point being equidistant 
from and D : a weight of 24 lbs. acting at N will produce the 
same effect as that which is a x 
produced by the two weights 
acting at C and D, and it 
may therefore be substituted 
for them. 

Then, if x be the force acting 
upwards at Ay 

X : 24=^JV : AB 



= 1:2; 
.*. 2x=24,.'. x=121bs. 



N 



Fia54 



B 
TV 



'12ll>S 



22 



KEY TO ELEMENTARY STATICS. 



16. (1) Draw ON at right angles to BQ, 

ThenO^:OJB=V3:2; 




Fia. 55. 



.-. ON- 



5^3 
2 



And P:Q=OiV:0^ 
or4:Q=.^:3; 



8^3 



lbs. 




Then since triangles AOMy BON are 
r similar, 



(2) Draw OM, OiV at right angles to AP, BQ. 

A , 

ONiOM^OBiOA 
=5:3. 
p X NowP:C=OiV:OM; 



Fia 58. 



/. 3 : C=5 : 3, and .-. Q=\\ lbs. 



(3) Draw OM, ON at right angles to AP, BQ. 

AO 3 




Fio. 57. 



Then 0M» 



V2 ;72 



••^•'^-2 V2' V2 2 ' 

- 2x3x2 2 X 3 X 2 X V2 ^g ii^ 
•■• '^"" 6, X V2 " 6l^T " 6 



KEY TO ELEMENTARY STATICS. 



EXAMPLB— VI. (p. 661. 



Then if C be the centre of gnvit^ 



2. Tha weight of 4 lt». at j1 and % lb«. at ^, the middle point t£ 
AB, are equiTalent to a w^ght of 6 Ibe. 

at D, AD being one-third of Alf, t 
Sinchea. 

.'. DB=10 inchea, and tha centre of 
gravity of 6 lbs. at i> and 1 lb. at B ia 
f of 10 inches, or 1^ inches from D. 

Hence the centre of gisTit; of the i\hs 
system IB 3^ inches from A. 

3. By Art. 71, if ^ be the centre of graTity, and D be taken 
the fixed point whose distance from . - f. (. 
Ifwe want to find in in<^iee. '~~ 

„^ P.AD*F.BD*P.CD 




4. To find the distance of If the centie of 
gtavi^ from A in feet. 

.„ S.ABi-iA0+2AD+l.AE 
^^~ 16+8+4+2+1 

8 + 8 + 6 + 4 26, . 



mn 

S 8 4 £ 1 



24 KEY TO ELEMENTAR V STATICS, 



5. The centre of gravity of B and C lies in 
BCy and it also lies in the line AD passing 
through the centre of gravity of the triangle. 

Hence it lies at D, the middle point of EC. 

.\ B^Cy and similarly it may be shown that 
*^A^B^a 





6. Let BG be the base, ABC^ DBC any two of the triangles. 
Draw A My DM to M the middle point of BC. 

Let be the centre of gravity of ABC, 
and draw OP parallel to AD, cutting DM 

Then shall iV be the centre of gravity of 

'b \ /mX \ P DBG. 

For since ON is parallel to AD, 
DN:NM=AO: OM (Euclid, VI. ii.) 
= 2:1. 



7. DEy joining the middle points of AB, ACyia parallel to BC. 

Hence AF bisects DE in 0. 
.*. the centre of gravity of DFE lies in FO, 
that is, in AF, which passes through the 
centre of gravity of ABG, Also, the distance 
of the centre of gravity of DFE from F is 
f . FO=i . AFy which is the distance of the 
K* centre of gravity of ABG from F. Thus the 
triangles have the same centre of gravity. 

Fig. 64. 



8. Taking the diagram of Example 6, suppose and ^ to be the 
centres of gravity of two of the triangles. 
Then since A0^20My and DN-=^%NMy 

.\AO:OM=DN:NM; 
.*. AD is parallel to ON, 




KE Y TO ELEMENTAR V STATICS. 



25 



9. Let P, Q be two equal particles ; P', ^, the same partides in 
new positions. 

Now, since FF'^QQf, and the angles of 
the triangles FF'Oy QQ^O are equal, each 
to each, / y^o 
.-. PO^QOy and rO^Q'O ; 

. *. is the centre of gravity of the particles D Q. Q' c 
in both positions. F,o. ^ 




10. Let D be the centre of gravity of A and By and E that of A 
and C. Let CD, BE intersect in 0. A 

Then the centre of gravity of A, B, C lies 
in CD and also in BE, 

.'. is the centre of gravity of Ay B, C. 

Hence JY, the point in BC where AO 
meets BCy must be the centre of gravity of 
B and C B N C 

Fio. 66. 




11. Let be the centre of gravity of the equi- 
lateral triangle ABC, Then AD bisects BC at 
right angles. 

Hence OB^OC; 

and similarly it may be shown that OA = OC, ^ 

.'. is the centre of the circle. 




Pio. 67. 



12. Taking the diagram of Example 11, 

J5Z)= CDy because is the centre of gravity of the triangle, 
and 0B= OC, because is the centre of the circle. 
.-. A ODB= L ODC^a, right angle. 
.*. AB=ACy and similarly it may be shown that AB—BC. 

,\ ABC is equilateral. 

13. To keep the line, passing vertically through the common centre 
of gravity of his own body and the weight, within the base covered 
by his feet. 



26 



KEY TO ELEMENTARY STATICS, 



D 
"7^ 



B 



3/6s 



Fio. 68. 



%[hi 




14. Let AB be the rod, its 
middle point, D the edge of the table. 
Then we have 3 lbs. acting at O, 
and 4 lbs. at B, 

Hence, since OB =7 inches, 
0Z)= 4 inches ; 
and .*. ^i>=ll inches. 

15. Let the weight of each particle 
be P. 

Then P at -4 and P at £ are equiva- 
lent to 2P at M, the middle point of AB. 

Also, P at and P at i> are equivalent 
to 2P at 0, the middle point at CD. 

.'. the centre of gravity lies midway be- 
tween and M. 



16. Let A be the point of suspension. 

Then the vertical line AD passes through the 
middle point of BG, and since BC is horizontal, 
the angles at D are right angles. 

Hence AB=AC. (Euclid, I. rv.) 



17. Let 0, P be the centres of gravity, A the middle point of th« 
common side, C the centre of gravity of the two. 
OA will be perpendicular to the side of the square. 
Draw BD at right angles to the same side. 
Let a be the side of the square, b the altitude of the triangle. 

ah 



Fio. 69. 





Then area of triangle »»• 
Area of square s^a^ 
.\^:a^=OG:BC 



~ OA : BD (by similar triangles) 



a 




Fio. 7L 



"2 '• 3 ' 



KE V TO ELEMENTAR V STA TICS, 



27 



18. ABC, A'BfC are the two positions of the triangle, CDBU 
being a vertical line bisecting AB and CA', 
Let iDCA=e, AC^Xy AB^y, 
Then z CAD^By and z ^DJ5'=2d, 
/. z ^'J5'J[>'= z ilFi>=»90'- 2d, 
and z ^^'2/=- z 5ilO=d ; 
.-. z^'J[>'B' = 180°-(90*-2^-d=90* + d. 



'^^^ 8in(90° - 2d)~A'iy~ x ' 



8in(90° - 26) 
co6$ _2y 



or. 



1/ 



cos2d 



X 



or 



apy ^2y. 






-1 



^2y 



'2x2-y« a; ' 

•. fl5*=4a;'-2y'; 
.-. 2y«=3»« ; 

/. x:y=V2 :V3; 
/. JC:CB«V2:1. 




Fio. 72. 



19. Place 3 lbs. at A, 4 lbs. at By and 5 lbs. at C. 

These are equivalent to 3 lbs. at Ay By C, together with 1 lb. at B, 
and 2 lbs. at C. /^ 

These are equivalent to 9 lbs. at 0, the 
centre of gravity of the triangle, together with 
3 lbs. at Dy one-third of the distance of C 
froni^. 

Hence the centre of gravity is one-fourth of 
the distance of D from 0, measured from 0. B DC 




20. Place 2P at B, 2P at 0, and P at A. 

These are equivalent to 4P at Dy the middle 
point of BCy and F at A, 

Hence the centre of gravity lies in AD at a 
distance from D one-fifth of AD, 




28 



KEY TO ELEMENTARY STATICS. 




21. Let A be the point of suspension. 

The centre of gravity of the parallelogram 

being the point of intersection of the diago- 

pnals, JC is a vertical line, and since the 

diagonals of a rhombus intersect at right 

angles ; 

.*. BD is honzontaL 



22. Draw MON^ BOS through 0, the point of intersection of the 
diagonals, parallel to the sides of the original parallelogram. 

Then MONy BOS wiU bisect the sides. 
Hence the centres of gravity, A, J5, C, D, 
of the four triangles into which the paral- 
lelogram is divided by its diagonals (all 
of which triangles are equal), are equi- 
distant from 0. Now the lines joining 
the extremities of the straight lines AC, 
BT>f which bisect each other, form a parallelogram, as can be easily 
shown. 





Pio. 77 



23. Let ADCE be a vertical line. 
'i MNC= L NAC+ACN 
^iCAD+ lA'CE. 
Now since AD bisects BCi 
.-. DC^^BC^AC; 

And since CE bisects A'B^, E is the 
centre of the circle described round 
A A'BfC; 

.-. ECr=-EA'. 
And .-. z A'CE^' l at ^'= z A ; 

.-. zMyC?'=46'+A 



KE V TO ELEMENT A R Y STATICS. 



29 




24 Let C be the right angle, CD the vertical line, bisecting AB 
inD. C 

Then z CDB^ i DCA+ i CAD 

= 2 z DAC (for DA=DB^DC). 
But z DAC+ L CBA^W ; 
or, zD^O+6zD^C=90% 
and.-. zi>-4 0=16°; 

.-. zCDB=30^ A 

Pio. 78. 

25. First remark that under the conditions of the question, if AB^ 
AC be the equal sides of the 
triangle, and the centre of 
gravity, z OBC^ z 0CB=46", 
since OD=BD=DC, 

Next, let BMCN be the 
vertical line passing through 
the points of suspension, and 
let B^C produced meet BC, or 
BC produced, in JR. 

Then ^ 

180°- z CJB(7= z CBM+ z RCB 
= z C5M+ z ^(TJ^ 
=:46• + 46•=90^ Fio.7d. 




26. Since AC=2BC, C is the centre A 
of gravity of the weights acting at A 
and B, 

,\ C m the centre of gravity of the y 
three weights. 




ilb 



3dts Zlbs 



Fio. 81. 



27. Let P be the weight of each of the particles acting at A, B, D. 



Then P at B and P at D are A 
equivalent to 2P at 0. 

Then if G^ be the centre of gravity 
ofPat^,and2Pat 0, 

AG='^ of AC=^^ o{ AD. 



B G 



Fia 81 



i 



30 



KEY TO ELEMENTARY STATICS. 




P 28. The limit in length of AB is attained 
when the diagonal ^C is perpendicular to 
BC, for then the centre of gravity lies just 
over C. 
In this case, since z ^1BC=60*, 

^-B= 250= 12 inches. 

29. BE bisects OC in K 

Then the forces, each equal P, acting at 
0, Ay By Cy DyEaxe equivalent to 2P at O, 
2P at Ny P at 0, P at 0, and therefore to 2P 
at 0, and 4P at N. 

Hence if G^ be the centre of gravity, it lies 
in OGy and is such that 

0(?=|- of Oiyr=y of OC. 

30. The centre of gravity of the body cpincides with the centre 
of gravity of the two parts taken together, and as this must be 
in the line joining Gi and G^, it follows that G must be in that line. 

31. is the centre of gravity of the square, G of the triangle, H of 
the figure AECD. Let 2a be the side of the square, and let EB=^x. 

0, ,C Then 4a^ - ox ; ax=OG : HO 

^MNiEM 
2 X 
=«-3-T^^"^' 

.-. 4ahi-aoi^-4a^ + ah:=ahi- -j-, and hence 

x=3a±a^3. 

Thus5^=aV3 ( V3- l),and ^JS?=a ( V3- 1), 
md.\BE:AE= ^3: 1. 

32. Let Af, i\r be the centres of gravity 
D of the equal triangles ABD, BCD. 

Then since equal weights act at M and 
Ny we may find the centre of gravity of 
the trapezium by taking the middle point 
Fio. 8$. of MK 





KE V TO ELEMENTAR V STATICS. 



31 




33. Let be the centre of gravity of the square, N the centre of 
gravity of the triangle, 2a the length of a side of the square, x the 
distance of A from 0. * 

Then area of square =4a', 

area of triangle ^^a {a + x), 
and AO: ON==a (a + sc) :4a*-a (a + x) ; 

or,a:a-^ -=a' + <w; :Sa^~ax; 

or, 3fl5 : 2a — a;=a + a; : 3a~» ; 
or, 9aac - 3a^=2a^ + ax - sc*. 

/. «* - 4aa;= - a', v p,o. 87. 

and a;=2a±V3a. 

34. Let J-B be the side on which the triangle will not stand, the 
centre of gravity of the triangle, D the middle point of AG, 

Now z ABC is an obtuse angle, 
.'. ACia the greatest side. 
And '.• z CDB is an obtuse angle ; ^- 

.-. CB is greater than CD ; ^^^ 

/, CB is greater than AB. 
Also, •/ DBA is an obtuse angle ; ^ 
.*. AD is greater than AB, 
,\ AB is less than half of AC 

35. Draw DE^ joining the middle points of AB, AC, 
Then triangle ADE is one-fourth of triangle ABC, 

Let jR be the centre of gravity required, the centre of gravity of 
triangle ADE, G the centre of gravity of triangle ABC. 

Then -^ X 0(?=-| X 22Gf ; ^ 

,\ Oa^ZBG, 
,\Aa-'AO^ZAB-'ZAa] 

.', 3^iJ=^^, or, AR=^ oiAF, 

36. D, the middle point of BC, is equi- 
distant from Ay By (7, if BAC be a right 
angle. 

.*., if be the centre of gravity, 

A0=^^ of AD'' J oiBD^% incbes. 





32 



KEY TO ELEMENTARY STATICS, 



37. Let BQC be the equilateral triangle. The limit of equilibrium 
is reached when the centre of gravity of the two figures lies in the 
vertical line BG, 



B 







M 


N 0.,^,..^-'^^ 



Fio. 91. 



Let 2a be the side of the 
square, x the greatest height of 
the triangle, Af, the centres 
^ of gravity of the two figures, N 
the centre of gravity of the two 
together. 

Then 4a« . MN-^ax. NO. 



X 



/. 4aS=aa; x --, or, 12a2=:a;«, and .-. x^2^3,a. 



38. Let 0, E be the centres of gravity of the figures, G the com- 
mon centre of gravity. 

Then 4a^. OG^ ^3.a\GE; 
or, 4 . 06?= V3 . {OE- OG) ; 

■"^ > or(4+ V3.)0(?= V3.(a + -i-.V3.a) 

=a(V3 + l): 




Fia 92. 



39. The triangle ABE is one-third of the quadrilateral DBCE. 
Let ^ be the centre of gravity of DBCE^ M that of ADE, that 
of ABC. 

.'. OM=Z.ON; 

or, AO-AM'^S. ON; 

or, jAF'-^AF^SON; 

.'. ON^^AF. 

.: NF=^AF- ^AF= ^.AF; 

.-.NF^^.HF. 




£^£y TO ELEMENTARY STATICS. 



33 












4 


37 


4 


fl 









40. (1) The given Bystem is eqniTalent to the system npresentoi 
in this diagram, if we combine eqmil parts of the 
forces, acting at the centrea of tiie outside squares 
on the light and left, in the centre of each middle 

And this system is equivident to the system repre- 
sented in this second diagtam, if we combine the forees 
Tertically. 

.'. the centre of grant; is the middle point of the 
middle square ; that is, the point of intersection of \ 
the diagonals. 

(S) The new system is equivalent to the ajstem represented in 
fignre 96, if we combine the forces horizontally. 

And this is equivalent to the ^stem represented in ' 
figure 97. 

Hence, if we take the struct line joining the 
centrea of the sqnaies marked 9 and I in the origins] 
diagram (since we have forces in the ratio of 3 : 1 act- 
ing at the middle point and the lowest point of this 
line), the centre of gravity of the Ejstem will divide 
the line joining the centres of 9 and 1 in the ratio 
5:4. 

41- Combining horizontally, the system is equivalent 
9i at if, 6^ at 0, 
12 at £^, 5 at j; 
&i at D, 3| at £. 

Hence if R and Q be the position of the points 
, -where tiiese forces, taken vertically, may be col- 
lected, a. b^ng a side of the sqoare. 



z 


4. 








It 


* 


2 


13 1 












0. 


^ 








= 






^« 



fl2 



HB— 



3 33a 11a 



H M 




R 




(4 

— — 


1 








3 





34 



KEY TO ELEMENTARY STATICS. 



Also, since 0, the centre of gravity of 30 lbs. at R and 15 lbs. at 
Q, is at the distance from i2 of | i^Q, 



ON^Ua^^Ha^^a, 



n-KT nrk , ^ / tt-d />in\ 41a 2 /11a 41a\ 14a 



42. Let a be a side of the large square, N the centre of gravity of 
the large square, It the centre of gravity of the portion removed, M 
the centre of gravity of the remainder. 

■^ Then MAT x(a«-^)=i2JVx~^; 

.'. MJVx 6a2=jBJVx 4a«. 





y^ 


/ 





Fio. 99. 



.-. MN^^MC, and .'. CM =i^, of CG, 
15 ' 30 



43. Let F and B be the middle points of J-B, BC ; D the centre of 
gravity of the straight bar ; G the centre of gravity of the bent bar. 
Then FB^4 feet, and BB=Z feet, and .*. FE^h feet. 

Then J^(?=y of FE = y ^®®*> ^^ -F!D=3 feet. 

a2 + 62-c* 



Take the formula cosC=- 



2a& 



D B 




Then coaDFG= 



(Trigonometry, Art 179.) 

FD^+FG^-Giy^ 
2FD . FG ' 
225 



or, -=- = 



9 + ^-GD* 



Pio. 100. 



_4 
5 



or, — = 



whence GD= 



2x3xy 

441+225--49gD2 
2x3x15x7 ' 
9V2 

7 



KEY TO ELEMENTARY STATICS, 



35 



,C 




44. 0, the centre of grayity of the complete hexagon, is equidistant 
from the angular points. Let the side omitted be indicated by 
the dotted line, and let the distance between 
M the middle point of this line and the opposite 
side be a. 

Then if G^ be the centre of gravity of the 
mutilated figure, 

OOx5=03fxl; 

"^^ 6 10 

45. The centre of gravity of a triangle coincides with the centre of 
gravity of three equal weights placed at the angular points. Hence the 
weight sustained by each prop is one-third of the weight of the triangle. 

46. Draw a radius CD from any point, and place the weight of 5 lbs. 





M 

Fio. 101. 




at a Produce CD to E, so that ^2>=f . CD. 
Draw the chord AEB at right angles to CE, 
and place the weights of 3 lbs. each at A 
and B, These are equivalent to 6 lbs. aoting 
at E, and 6 lbs. at Ewill balance 5 lbs. at CE, 
because DE : 0D=5 : 6. p,o. 102. 

47. Let the line drawn parallel to the base cut the line ADy joining 
the vertex of ABC to i>, the middle point of BC, into two parts, 
m and n, of which m is the part nearer to the vertex. 
Then area ofAEF : area of ABC =m^ : (m + n)^ 

Euclid, VI. xix. 
Let and G be the centres of gravity of the 
two triangles, and B the centre of gravity of 
the quadrilateral EBCF, 
Then GBx{{m+ny-m^^OGxm^. 
i jn 2(m + n) j ,^ „> 2 „ 

^ o } 6 Pio. 103. 

. -P_2 { (TO+»)»-m»} 




^ 



36 



KEY TO ELEMENTARY STATICS. 




48. Let A be the point of suspension. Then, 
by the conditions of the question, AD bisects BC 
at right angles. Hence AB=^AC (Euclu), L iv.) ; 
and .*. the four sides are all equaL 




49. Place w at Ay 2w at B, 3io at C w at 
A and 2w at B are equivalent to Zw at JD, 
AD being equal to twice DB. 

Join DC, and bisect it at G, then G is the 
centre of gravity required. 

AhoAG:GE=6:l 
CG:GD=^l:l 
BG : GF=-2 : 1. 




60. Let G be the centre of gravity of the Utnin^ ^ 

and D the middle point of AE the perpendicular on 

BC. Let X be the weight placed at A^ 

ThenxxAD=^6xDG; 

AE ^ [AE AE\ 
or,xx — =6x(^— -_j; 



or 



6^ 
6 



^|.=Z^and.-. a;=21bs. 





Fio. lor. 



51. Let A be the point of suspension. Then 
AD is perpendicular to BC Let P and Q be the 
weights. 
Q Then P:Q=aD:5i>. 

lSowBD:BA^BA:BC} _ ^ 

and CD : C^ = C^ : 50 1 ^^^°' ^- ^^• 

.*. 5D= jg^ and CD= -g^ ; 

/.FiQ^CA^iBA^. 



KEY TO ELEMENTARY STATICS. 



63. At the limit of equilibrium AC is vertic 

.-. the iucliniktioo of the ptane is 45°. 

If the incUnation of the plane be greater 

than 45°, the centre of grayity of the block 

will be outside the base, and the block will 



63. ^ is the centre of gravity of the two 
smaller squares, and H is the cenire of gravity 
of the sqnaie on the hypotenuse. 

Now the weight of the two squares collected 
at ^ is equal to the weight of the larger square 
collected at D, and therefore the common 
centre of gravity is at 0, which bisects AH, 
and also bisects the hypotenuse of the triangle. 




Examples— VII. (p. 87.) 

1. Let AT>—x feet. 
Then I>£-(S-3:) feet, 

andic-e (S-i), .-. a:=3. 
Pressure on the fulcrum =4 ]hs + 6 lbs = 10 lbs. 4- 

2. DB:AJ)=1:^. I 

P+Q=361b3. 
NowP:g = DB:JD; 
.■. P:36-P=7:9, .-. 9P=352-7P; 

.-. P=15ilbs.,aode=20ilhs. P p^j^, 

3. Let X be the weight of the rod, which may be supposed b 
at M, the middle point of the rod, ' ' " 



Thenx:3-DJ7:Mi> 
=4:3J; 

.•.ia!=12,or,a;=V=3?Ilw. 



38 



KE Y TO ELEMENTAR Y ST A TICS. 



A6 



A D 



B 







Fio. 118. 



4. Let AD^x inches. 

Then 6^D + 6CD=10DB, 
or, 6a; + 6 (10-a;)»10 (7-a;), 
or, 6a; + 60-6a;=70-10a;; 
.'. xsl inch. 

Pressure on fulcnim = (10 + 6 - 6) lbs. 
» 10 lbs. 



so 



w 
Fig. 114. 



5. Regard B, where the rod rests on the shoulder of one man, as a 

fulcrum, and let x be the pressure on the 

A C B shoulder of the other man ; w the weight 

^ of the rod, acting at its centre of gravity, 
distant from B one-third of the length of 
the rod. 
Thenx:ir=^C:^^ 
= 1:3 
.'. 05= one-third of w. 
Hence pressure on ^ = two-thirds of w. 

6. Let AB be the original length of the lever, D the fulcrum, 
DB = one-third of AB. 

Q_ Then C=2P. 

Now let Q act vertically upwards 
at By and let C be the new position 
4? of the fulcrum. 

Then WC=AC 



B 



Tia. 115. 



^AB+BC; 
.-. BC=AB=^3DB. 



vv 



\ 



7. 3f is the middle point of the lever. 

A D_M Let ^2>=x feet. 

Thenl0x=50(3-a;), 
10x=150-50r, 
and.'. 2=2J^feet. 
Also pressure on the fulcrum =60 lbs. 



10 



60 
VI0.11& 



KEY TO ELEMENTARY STATICS, 



39 



8. Collect 1/7, the weight of a 
the beam at the point N round 
which it balances. 

Then w x DiV=100 x DB, 

or, ii;x8 = 100x2; 

/. IT = 25 lbs. 



Ji 



11/ 



Fio. 117. 



D 



B 



JOO 



9. M is the middle point of the rod, 2t 
w is the weight of the rod. 

Then 2 J x ^D = w x 3fD + 2| x BZ>, 
or, 2J X 4=ir x 2 + 2| x 1 J, 
or, 10=s2w + 4, and .*. w=^ lbs. 

Also, 
pressure on D= (3 + 2§ - 2J) lbs. =3J lbs. 



M B 



w 24 

Fig. 118. 



D 
TV 



10. Regard the peg D as a ful- 
crum, and let x be the pressure on 
the peg C. 

Then SxAD=x>iCD + 6xBD, A 

or, 3x5=4« + 5xl ; 

.'. a=»2J lbs. 
Hence 
pressure on D= (8 - 2^) lbs. = 5 J lbs. 



Next, let 2p be the upward 
pressure applied at N, a point 
2 feet from B, 



X 



AT 






Then pressure on each peg 

=i(8-2|))=4-p. 

Make D the fulcrum, as before. ^ [ 



7\ 



Fig. 119. 



A. 



B 



N 



Fio. 120. 



Then 3 x AD={^-p) x DC+2p x ^2) + 5 x DB. 
15=16-4p + 2p + 5, .-. 2;>=61bs. 



B 



KEY TO ELEMENTARY STATICS. 



11. Let DB-x iocbet. 

or,2(17-a;) + 7(15-i) + B(13-jr)- 

or, 34-2x4- 105-71 + 65 -Si-ai; 

.-.a; -12. 



12. The adrantoge Till be increaaed, becatue the dititauce of the 
power from the fulcrum b increaaed. 

13. Let i£ be the middle point of the bar, and suppose the pressure 
on the prop it to be twice as great oa that on A, the prop that is 
1 foot from one end of the bar. 

^ Then pressure oaA— -»-• 

BegBjd B aa the fulcrum, and 
W 

make an upward force of -^ acting 

•if] at A repreaeut the reliance of A. 

kAB'W^MB 

- W. [AB-Am = f^- (JB-4) ; 
.-. AB~ZAB- 12, or, AB-Q feet. 



14.LetCbetliepointof 

application, CB = x inches. 
Then 



KEY TO ELEMENTARY STATICS. 



41 



15.xxOD=8x^D + 8x^J>, 

or, 16a; =8 (18 + 12); 

/. 2a; =30, or, x=16 lbs 

The force, being less than 

the sum of those which it 

balances, acts at a mechanical 

advantage. 






B 



1 




TV 



t 



Fig. 124. 



16. Make B the fulcrum, and re- l 
gard the resistance of A us an up- 
ward force »a; stone. M is the 

A 



M 



A 



3 

A 



i i 

Fio. 125. 



middle point of the beam. 

Then x x ^5=4 x 2)5+ 16 x MB, 

6a;=4x4 + 16x3; 

.*. a;=10f stone; 
and .*. pressure on B=(20- 10§) stone =9 J stone. 

17. Suppose the weight supported by the man at B to be p, and 
the weight supported by the man at ^i to be rp. 
Then the weight of the beam acting at the middle point Jf is (r + \)p. 



Regard B as the fulcrum. 
Then rp x ^5= (r + 1);> x MB, 
Let 2x be the length of the beam. 
Then rp. (2x- a- 6) = (r +!);>.(«- 6), 
or, r (2x— a-6)=ra; + x-6r— 6, 
or, 2ra; - ra=ra; + a; - 6, 

or (r- l)x=»ra- 6, and .'. 2a;= -^^^— <* 
^ ' r-1 



Ay? 



M 



A^ 



^B 



V+flp. 
Fio. 126. 



18. Let M, .N* be the middle points of the axes of the cylinders, and 
the point where the axes meet. M Q |s| 



Now since If = f ft., and NO = « ft,S 
and since 15 x |=9 x |, 
it follows the cylinders balance at 
0, the point of junction. 



S 



3) 



iS 



"E\<i.vn. 



42 



KEY TO ELEMENTARY STATICS. 



15 



BD 



130 

Pio. 128. 



25 



19. Let OZ>=a; feet. 
Then 15 X (12 -a) + 130 (6-a;)=25a;; 
or, 180- 16a; + 780- 130a; = 26a:; 
or, 960 = 170a;; 



20. Let D be the point at which Wy a weight equal to the weight 
of the rod, is placed. 

Let M be the middle point of the rod. 



f 



M 



T 
tlT 



T 
Fio. 129. 



4w 



Then pressure on^ =§ ot{w + «<;)=- 

o 

Regard B as the fulcrum, and repre- 

■J^ sent the length of -4 B by x. 



4W7 / ^ TXN X 

Then -5-xa;=i/7x (a;--4X>)+tr.— ; 

o Z 

4a; ^ T^ . « 



a; 



.*. ^D =—-—=—= one-sixth of the length of the rod. 
2 3 6 ^ 



X 



Fio. 130. 



152 



B 




21. Regard B as the fulcrum, 
and let a;, equal to the weight 
sustained by A, act upwards at 
A. 
Then x x J[j5= 152 x 2>j5, 
or XX 38=152x14; 
.*. x=56 lbs. ; 
and .'. pressure at JB=96 lbs. 




22. Let w be the weight of the beam, which may be supposed to 

A C M D act at Z>, roulld which the beam 

balanced in its first position. 
Then 200 x AM+ 20 x CM^^w x MD ; 
or, 200xl2 + 20x8=2ti;; 
Fio. 131. .'. 2t(;=2560, or, w=1280 lbs. 



200 so 



Y 
ur 



KEY TO ELEMENTARY STATICS. 



43 



23. Let M be the middle point, D the fdlcmm at a distance of x 
feet from the end N, 



N- 



A B 



•f y >r 



C E p M 



3 5 7^ iSO 

Fio. 182. 



Then 3^D + bBD + 7CD+ 9ED = 150MD. 

Or, 3(x-2) + 5(x-4) + 7(x-6) + 9(a;-8) = 150(10-a:) ; 
.-. 174x«1640 ; or, a;=^9|f feet 

24. (1) Let If be the middle point of the lever, W the weight of 
the lever, 2> the fixed point, B the end where P, when suspended 
from it, balances the lever. 



Then WxMD=PxDB, 

Let A be the point where P 

is placed to balance nP acting 

at B. 

Then 
PxAD+WxMD^nPxDB; 

OTy PxAD-k-PxDB=nPxDB; 

OTfAD + DB=nDB; 

.'. AD=(n-l)DB, 

(2) Here^D=10xi)j5; 

.'. 10=n-l ; or, n«»ll. 



M 



D 



W 
Fio. 13S. 



nP 



25. Let If be the middle point 
weight mnst be attached to somepoint 
in ^D,and that it may be as small as 
poesible,^ it most be made to act at A, 
Let X be the additional weight 

Then 

(x + 3) X ^2>=:2 X DAf 4-4 X DB ; 
or, x + 3 = 2+12, and.'. x=5ll lbs. 



It is clear that the additional 



M 



X-hS 



2 

Fio. 134. 



44 



KEY TO ELEMENTARY STATICS. 



26. Let W and 3 TT be the weight of the rods, JVand M the middle 
points of the rods, x the weight attached to W, D the fulcrum, A 

and B the points in the horizontal 
line through the fulcrum, through 
which the vertical lines through y 
and M pass. 

Then since z ADN = i DBN, 
and z DAN= l DBM ; 
. *. triangles ADN, BDM are similar. 
Then Pr+a;;3Tr=2)5 ^D 

=3.1; 
W+x^9W; 
/.x=SW. 




FiO. 185. 



27. Let M be the middle point of the rod, (7 the first position of 
the fulcrum, D the second position of the fulcrum, W the weighty 
and 2x the length of the rod. 

5* 



C D 



M 



B 



T 
Fio. 136. 



Then5f x(2a;-2) = Tr(a;-2), 

and6(2x-3) = (Tr+l)(x-3) 

Hence56x-66=5Trx-10Tr, ) 

lOx-lb^Wx-SW-^-x-d, ) 

OT,bex-b6 = bWx-lOW, ) 

bOx-7b=-bWx-lbW+6x-l6;) 
.-. 6x + 19 = 5Tr~6x+16, 
or, 5Tr=lla;+4. 
Substitute this value of STT in (3), and we get 

56a;-56=llx2_i8x-8, 
and from this we find a =6, and .'. 2x=12, and W=14, 



(1) 
(2) 

(3) 



KEY TO ELEMENTARY STATICS, 



45 



Then Tr(a;-2)=5x2 ) 
Tr(a;-l)=llxl ) 

Wx-W ^\\S 



7V7^ 



28. Let TT be the weight of ^ [^ q 
the rod, 2x the length of the 
rod, C and D the two positions 
of the fulcrum. 



M 



w 

Fro. 18T. 
W^=l, and.*. 2x=24. 



M 



TV 



29. Let Jlf be the middle point 
of^i;. 

Thenl2x AD=8xDB+xx dm/'- 
or, 12xll«8x4 + a:x5; 

.•. a;=28f lbs. 

f2 Fio 188. 

30. Let DB^x. 

Then9x7 + 7x6 + 6x3«3x2 + 6x4 + 7x8+x, 

D B 



Fro 180. 

or, 63 + 42 + 15 = 6 + 20 + 56 + «; 
.'. a;«38 feet. 



31. From D the fulcrum draw DB at right 
angles to the direction of x, the force acting at 
the longer arm. 

Then since z CDB-60*, 

/. CD-2DB. 

Kowa;xi>£«6x^ 

or, a; X j >-6 X 3, and .'.a^— 7J lbs. 



^/ 



8 




46 



KEY TO ELEMENTARY STATICS, 




Fiu. 141. 



32. From D the fulcrum draw DiV at 
right angles to the force of 10 lbs. 

Then AB : DiV= 2 : ^3, 
and 10xDiV=20x4; 

.-. DiV^=8, and /. -4D= ^ = —3-^ 
.'. the length of lever is — ^ feet 



33. One way clearly is to reverse the position of the rods, making 

the 1 lb. weight of the one correspond to the 9 lbs. weight of the other. 

» Q Another way is found in the following 

manner : — 
Let D be the fulcrum. 
Let X == the distance of the end of the 



\ t 



\j 1/ f Y ^' upper rod from the fulcrum. 
' ^ ^ Then 

Fio. 142. 



lxa; + 3(a;-l) + 6(a;-2)x7(a;-3) + lx24-3x 1 

= 9(4-a:) + 7xl+9x2, 

or, x + 3a;-3 + 5a;-10 + 7a;-21+2 + 3=36-9a; + 7 + 18, 

or, 16a; -29 = 61 -9a;, and .*. a;=3f feet ; 

.". the 1 lb. weight of the upper rod projects 1| feet beyond the 1 lb. 
weight of the lower. 



I 



34. Draw BD at right angles to the direction of the force of 4 lbs. 

Then 2 X ^£=4x51) 

^4xAB,amBAD; 
.-. sinJ5^D= J, and .-. z BAD ^30\ 
Also, since the pressure on the fulcrum 
acts along AB, if i2 be this pressure, 
42=222 + 2*, or, 2^2=12, 
Fjo. 243. and .'. i2=2 V3 lbs. 




KEY TO ELEMENTARY STATICS. 



47 



A B D 



35. Let FT and 2Tr, the weights of the beams, act at M and N 
their middle points. 
Let P be the weight required. 
Let ABDC be a horizontal line. 
Then PxAD+Wx BD=-2Wx DC. 
Now BD=ND, and /. AD-^DC, 
And RM^MD, and .-. BD=\AD, 
Hence Px ^D+ WxiAD^2Wx AD, 




Fio. 144. 



W 



3fV 



or, P+ 2-=2fr, and .*. P= ^ • 



36. Draw MEN At right angles to the directions 
of the weights, and BD vertical ^ 

Then since the weights are equal BM^BN; 

.;. AD^DC, (Euclid, VL it. Ex. 1.) M 

Hence, since ABC is a right angle ; 

,'. DB'^AD^DC. (Euclid, IIL XXXI.) p p 

.% z DAB= L ABD, and z DCB^ l DBC, fiq 145, 




37. Xet ADB be the lever and D the fulcrum, 
CDB a line perpendicular to JIP, PP, Jf the a 
middle point of ^iP, MD a vertical line. 

Then since AP, MD, BP are parallel ; 

•. CD : DE=AM : 3fP. 

(Euclid, VL II. Ex. 1.) I 

.*. CD=DE, and .'. there is equilibrium. 




Fio. 146. 



38. Let MAN be a horizontal line. Then since 
AB and AC sie equally inclined to the vertical, 
they are equally inclined to the horizontal ; 

^ .\ lMAB^lNAC', 

.*. MBA, NCA are similar triangles. 

l!heaP:Q=:^AN:AM 
^ACiAB 
*1 :2. 




irva,\« 



48 



KEY TO ELEMENTARY STATICS. 



39. Let D be the fulcram, MBN a horizontal line. 




Then i>lf = 



AD 



X 



Fio. 149. 



and DiV=^.D5; 

.•.xxi|=16x^xl2; 

16x^3x12x^2 16/i/6,v 
== 27T8 ^3 ^^' 

40. Let P and Q be the forces acting 
along AB and AD, 
Draw CAf, CN perpendicular to -4P, 

Then since z ^PC= z ^2X7 ; 

.-. iCBM=iCDN; 

.*. triangles BMC, DNC are similar ; 

.\F\Q^CN:GM 



U 



T 

f2 



Fio. 150. 



Examples — ^VIIL (p. 96.) 

^ \. W will be greatest when the 
1 moveable weight is at D, the extre- 
J mity of the steelyard. 
' Then ^GD x 12a(?= W. CA, 
or, ^x 33 + 12 xl4=5fr, 
and .-. fr-42f. 



W 



2. First, when G^ the centre of gravity is in the longer arm. 

Let P be resting at the graduation marked n. 

Then the steelyard says that W'^nP, or, W^2n lbs. 

BD GCOA BD OCGA 



W 



P 



Q W 



^ 



Fio. 161 
But really 2. CD + Q. C(7-Tr. CA (a), 

a,ndQ,CG=l.CO; 
.'. 2CD+C0'-'W. CA, or, 2 (OI)-CO) + CO> 



W.CA. 



KE Y TO ELEMENTAR V STATICS, 



49 



Also, OD^nCA ; 

.\2n.CA-C0=^W.CA; 

CO 

,\ Tr=»2n-7=-j, and /. W is less than 2n* 

Next, when lies in CA, the equation (a) becomes 

^D'=W.CA + CO, 

0T,2 (OD + CO)^W.CA + CO; 

.-. 2n.CA + C0= W.CA; 

CO 
,\ Wss2n+rj-2, and .*. W is greater than 2n. 



3. PxCD+ — xOGf-TTx^C. 

P 
Let acss length of rod. 

P3a; , P «. TIT- Jc 
4 p 4 4 



.-. 3P+^«fr, or, Tr=(^i-^y. 



W 



P 

Fio. 152. 



P 

4. There is no limit to the increase in the magnitude of P, where- 
ever the fdlcrom may be, provided that the graduations can be made 
soffidently small ; but there is a limit to the decrease in the magnitude 
of P, except when the fulcrum is at the centre of gravity, beoiuse P 
may be too small to balance Q when P is made to act at a point in 
CA. 

5, Let B be the weight suspended at Q to make the instrument 
correct for a moveable weight nP, 

TheaP.CD-i-Q.CG^W.AC, (1) 

AadnP.JDD+(Q+B).Ca^nW.AC, (2) 

G IC 



T 




Fia.lS8L 

Henoe, subtracting (1) from (2), 

(n-l).P.CD+B.CO^in-l).W.AC, 
OT,B.Ca^(n-l)(W.AC'P.CD) 

=(n-l).^.0(?; 

.\B^{n-l).Q, 
D 



from(l) 



50 KE Y TO ELEMENTAR V ST A TICS, 



6. If a be. the real weight, 

a; : 38 = 20 : 19, and /. x=40 lbs. 

7. The arms of the balance are as 14 : 16 ; 

.*. when the body is placed at the end of the longer arm, apparent 
weight : 16=16: 14; 

/. apparent weight = — rj— =18? ounces. 

8. Let X and y be the lengths of the arms, 'p and q the apparent 
weights. 

Thenjpa;=y, and qy=x; 

y X 
^ ^ X y' 
x^ + y^ 13 



••* g — y 9 

"^^'^ 6 ^ 144 "144' 
13y 5y 

or X — • — =s — ^ • 

^^'^ 12 12' 
••• |"=f or §. 

9. He gets 15 ounces for 3s. 9d. 

.*. for 16 ounces he pays — ^-— d., or, 48d., 
that is, he buys at the rate of 48. per lb. 

10. Take the diagram on page 95 of the Statics, 

P= l&n ounces. Let the graduations nm from to ^. 

When ir= 1 ounce, ?-^=~» and .'• OC=^:r^^. • AO. 

AC IGn 16/1 + 1 

When ir=2ounce8, -—=,-—, and.'. OC^r^ r -4 0, and so on. 

AC 16n' 16w + 2 



KE V TO ELEMENTAR V STA TICS. 5 1 



Examples — IX. (p. 99.) 

1. Let X be the radios of the axle in inches. 

Then3:18=x:3xl2; 

3x3x12^., 
.-. x= — r^ =6 inches. 

2. Let X be the power in lbs. 

Thenx:3=2:6; 
.*. 6x=6, and .*. x=l lb. 

3. Let X be the power in lbs. 

Then x : 12=3 : 9 ; 
. . 9x=36, and .'. x=4 lbs. 

4. The capstan is a strong cylinder of wood moveable about a 
vertical axis, with short bars, called handspikes, inserted near the top, 
by means of which it can be made to revolve. A rope coiled round 
the capstan is /ittached to the weight that has to be raised. Hence 
mechanical advantage is gained, because the distance of the ends of 
the handspikes from the axis is greater than the distance of the coil 
of rope from the axis. 

Let X be the weight the men can support in cwt 

Then6:x=2 :5 ; 
.'. 2x~30, or, x=15 cwt. 

5. If a vertical line be drawn through (see diagram on p. 98 of 
the Statics) and P be made to act at the end of this radius of the 
wheel, P will produce no pressure on the axle, aud therefore the 
pressure on the axle will in this case be the least possible. 



6. Since P and TV are equal, it is plain that 
equilibrium can only exist when the moments 
of P and W about are equal, and this will be 
when the direction of P is a tangent to the , 
axle^ as in the diagram. p ^ 

Fio. 154. 



@ 



52 



KEY TO ELEMENTARY STATICS. 



7. Let X be the greatest weight that can be supported. 

Then36:a;=l :3; 
/. a; =108 lbs. 




FiQ. 165. 



8. Let TFi and TFj be the two weights. 
Let be the real centre of the wheel, 
and A the axis of the axle. 

Then P x AM= W^ x AD, 
andPx^iV=Tr,x^^; 
.\P,{AM^-AN)^W^y^AB^W^xAD^ 
.-. 2P X 0M= (TTi + W^jAD. 
But if and A coincide, 

2PxOAf=2fr.^D; 



9. The greatest weight will be supported when W acts at the 
extremity of a diagonal, and the least when w acts along a side of the 
square. 




p : W^\ diagonal of square : radius of wheel, 
F :w^\ side of square : radius of wheel ; 
.'. TT: to = diagonal of square : side of square 
= V2 : 1. 




Examples— X (p. 105.) 

1. Let 26 be the angle between the strings. 

Then2P,cos^=Tr, 

and .'. 2P. cos^=P, and cos^= J, or, ^«60*, 

and .•.2^= 120*. 



KEY TO ELEMENTARY STATICS. 53 



2. If If? be the weight of the pulley, 

.*. if to be not less than P, 
W is not greater than P, and there can be no advantage. 

3. P: Tr=l:2cosd 

= 1 : 2cos30*=l : 2. 51^-1 : V3. 



4. P : 8=1 : 2», and /. P=l lb. 

5. Let X be the weight of each pulley. 

Thenll = V + |- + f+|, 
or, 88=32 + 7x, and /. a=8 lbs. 

6. P-f+f +f +f ; 

.-. 8P«6 + 12 + 6 + 3, and /. P=3J lbs. 

.-. P=10 + 2, or, P=12 lbs. 

8. P=|+i + i+i; 

.•.P=y=2, or,P=21bs. 

^ Ty W W W 

.-. 8P=7TF, or, P : TF=7 :8. 
10. Let X be the weight of each pulley in lbs 



XXX 



Then75-V^ + ^ + ^ + -f.; 
or, 63 =56 + 7a;, and .-. a=l lb. 

11. Let X be the weight of each pulley. 

ThenTr-,^+^ + i^+*-; 
8248 

or, 8Tr= W+ 7x, and .% ac— TT. 



< 



54 



KEY TO ELEMENTARY STATICS, 



12. Let the number of pulleys be n. 

Then W=v)+p, 

and;7=^^jj^, or, 2*.|>-j?=w; 

/. «;=(2**-l)jp, and 2"-l is necessarily an odd number, because 
all powers of 2 are even numbers. 

13. 2=|^+j+i+J+iV; 

or, 32= W+ 8 + 4 + 2+ 1, and .-. W= 17 lbs. 

14. Pressure on beam=F + 21^ + 4P'+ . . . 2«-i . P' 

=(2'»-l)F; 
/. Tr'=(2»-l)P'; 
also, W=2^.P; 

.••5^-5^-2--(2«-l)=l. 

15. Let D be the fixed point, 
p The beam is acted on by two parallel 
forces, 

one in BM=P, 
one in CN=-2P; 
,\ there is equilibrium when 
BD:CD=^:P 
=2:1. 

16. P:8 + 112=1:G; 

.-. 6P=120, and .'. P=20 lbs. 

17. 2:TF+8=1 :7; 

.-. 14= W+ 8, or, W= 6 lbs. 

18. Let W be the weight of the man, P the power he exerts. 

W 
ThenP: -2- = l:7; 

W 
• p=— . 

•• -^ 14 
Hence pressure on noor= ^ " TJ == IJ"' 





KEY TO ELEMENTARY STATICS. 



55 



19. P:TF+3P=1:6, 

or, (iF= W+3P, and /. TF=3P 

20. Let w be the weight of the lower block. 

Then3:10 + ir=l :4, 
or, 12«lO+t0, and .*. w=2 lbs. 

21. Let n be the number of pulleys aD the lower block. 

If the string be fastened at the upper block, the number of strings 
at the lower block is 2n,. 

If the string be fastened at the lower block, the number of strings 
at the lower block is 29i + 1. 

In the first case, p : w+p=l : 2n, and .'. w= (2n— l)p. 

In the second case, jp : w +jp = 1 : 2» + 1, and .". w^^np. 



•K 



22. Let D be the point of suspension. 

Then, referring to the diagram on page 104 of the Statics^ the action 
of the strings on the bar will be represented by ,p gp p 
three equidistant vertical forces, P, 2P, 4P. 

Let 2a be the length of the bar, x the distance 
of D fipom the force 4P. 

When there is equiUbrium, the bar being ~ 
horizontal, taking moments round D, 

4P X x=2P X (a-jr) + P X (2a-a;), 
or, 4a;=2a-2a;+2a-a;; W 

.•. 7a; = 4a, or, a; = fa. Pio. iss. 



23. Taking the diagram on page 104 of the Statics, extended to 
six pulleys, but omitting the string RA, and taking w for the weight 
of each pulley, 

tension of first string, SB =w, 
tension of second string, TC=2w^ 
tension of third string =7w, 

tension of fourth string = 15m7, 
tension of fifth string = Zlw ; 

/. w+Zw + 'Jw + lbw + Zl'tr^W I 
.-.w: W^=l :57. 



56 KEY TO ELEMENTARY STATICS. 



% 



Examples — XL (p. 112.) 

1. Since 62 + 122=169=132. 

F : IF= height of plane : leogth of plane 
=5:13; 

.'. P= -rr- tons=l^ tons. 

2. Since 32 + 42=25=52, 

W : 12= length of plane : base of plane 
=6:4; 

... U=12;^lbs.=81bs, 



3. W will be represented by 6 ; 

/. ie=}fF, andP=fTr. 

« 

4 The height of plane will be represented by 3. 

P : Pr= height of plane : base of plano^ 
P: 12=3:4; 

.-. P=af lbs.=91bs. 

6. P : P= height of plane : length of plane ; 

.'. 1 : 2=height : length ; 
.'.1:2: iv^3= height : length : base. 
Then P : fr=height : base 

.-. TF=V3 lbs. ; 

and if ^ be the inclination of the plane, 

sin^= J, or, ^=30^ 

6. Let P and P' be the two forces, 

height : base=P : W 

= 15:20=3:4; 
.*. height : base : length =3 : 4 : 5. 
Hence P':20=3 :6; 

.-. P'=12 IbB. 



KE Y TO ELEMENTAR Y ST A TICS, 5 7 

7. The resistance to motion arising from the rails is (50 x 10) lbs., 
and therefore the tension of the rope is thus relieved to the extent of 
600lb8. 

Also P: 50=1 :30; 

.'. P= J8 tons ; 
/. strain on rope =(jj -5*55%) tons=l§}f toms. 

8. P : }F» height of plane : base of plane, 

and .*. if a be the inclination of the plane, 

tano=^=r-^=V3> a^d •*• «=60*. 

9. P : fTss height of plane : length of plane ; 

/. P: 2^2=1 :V2; 
.-. P=2 lbs. 

10. P : Tf™ height of plane : base of plane ; 

.-.P: 56=1:1; 
.-. P=56 lbs. 
Hence any force greater than 56 lbs. will move the weight. 

11. P : TFb height of plane : base of plane ; 

.-. P:12=V3:1; 
/. P=12V31bs. 

12. P : W^ height of plane : base of plane, 

and the angle of inclination is 30*^ ; 

.-. P:10=l :V3; 

. p 10 „ 10^3,, 
.*. F—'-pz lbs. =s — ^ lbs. 

13. Let P and 2P be the forces. 

Then -^ ^heig ht of plane 
W length of plane' 
^^ 2P^height_of_plMie . 
W base of ^lane ' 
. 1 __ base of plane 
length of plane * 
,% angle of inclination is 60^. 



58 



KEY TO ELEMENTARY STATICS. 



14. Let P be the tension of the string A W, 

A Then P : W^= height : length 

W 




.-. P= 



72* 
Hence pressure on the ground 



Pro. 159. 



=Tr- 



V2 



-'KS^)-<-^') 



15. P : fr=height : length ; 

/. P:4=«l:2, or, P-21bs. 




W 
Fig. 160. 



P 16. Let I POD^ L ROD. 

Draw DM parallel to OF, 

Then the sides of MBO are parallel to P, W, R, 
Also z 3fD0= z DOP 

= z DOM. 

Now P:i2=DM:0M. 

But DM^OMy and .-. P=JK. 



17. Let Pi be the power acting parallel to the plane, Pa the power 
acting horizontallj. 

Then Pi : TT^ height : length, 

and Pf : TF= height : base ; 

.*. Pi X length = Pa X base. 

Now, the length is greater than the base ; 

.*. Pi is less than Pg. 



Ik 



18. The string will just support the weight when 

height of plane : length of plane =10 : 20 

= 1:2, 
and in this case the angle of inclination is 30^ 
.'. as soon as the inclination is greater than 30**, the height of the 
plane is increased, a power greater than 10 lbs. is required, and the 
string breaks. 



KEY TO ELEMENTARY STATICS, 



59 



19. The tensions of the parts of the string A W and A W are the 



same^ represent each by P. 

.W AG 
*"^p-=Xi>' 
. W AB 
"W^AC' 



j/J^ 



B DC 

Pio. 161. 



20. The power acts first along the plane and then horizontally, and 
taking 2a and a as the angles of inclination, and 27rand 3Was the 
weights supported in each case. 

;.= tana ; 



.*. 28in2a=3tana; 
3sina 



.'. 4sina.cosai 



cosa 



.*. cos*a=-^-, or, cosa=^ ; 
.*. 0=30", and .*. 2a=60*. 



21. Let a be the inclination of the plane. 

10 height 
^**^Tr=leigth^ 

, 20 height 

*°*^TF==-bSi"^ 

._b^e_ ,, 

"length ^^ ^' 

.*. a is an angle of 60*. 



22. Height of plane : length of plane = 12 : 20=3 : 5 ; 

.'. height : length : base = 3:5:4. 
Naw, horizontal force : 20=3 : 4 ; 

/. horizontal force = 15 lbs., 
and 15 : 12=5 :4 



i 



6o 



KEY TO ELEMENTARY STATICS. 



Again — 

Pressure in first case : 20=4 : 5 ; 

.*. pressure in first case =16 lbs. 
Pressure in second case : 20=5 : 4 ; 

.'. pressure in second case = 25 lbs., 
and 25: 16=52; 42. 



23. Height : length : base =1:2: V3, 

and hence the angle of inclination is 30' 



24. ^C: ^0=3:5; 

Hence if AB become the height of the plane, 

3 lbs. : weight supported = 4:5; 
.'. weight supported = 3| lbs. 



25. The force required to keep W at rest, acting along the plane, 
is W , sina. 

Hence the force acting down the plane must be W . tana - W , sina, 
for then the resultant force tending to pull the body upwards along 
the plane will be 

TTtana- (TF". tana— PT. 8ina) = TF. sino. 



26. There will be equilibrium when P and W make equal angles 
with the direction of 12, each angle being evidently a, the angle of 
inclination of the plane. 



7 


Y\ 


sm2a sma 
and .*. Bi=2W. cosa. 


^c 




But when P acts parallel to the plane. 


y 


f 


B=W.coaa. 


\ 
Fia.: 


V 
162. 


Hence Bi=2B. 



KEY TO ELEMENTARY STATICS. 6i 




Miscellaneous Exercises (p. 121). 

1. 0-4, ^the direction of the re- 
sultant of the equal forces P and Q, 
bisects the angle TO(i, 

OA is opposite the resultant of the ^ < ^iy^ > A 

equal forces R and iSf, and if OB be 
tihe direction of this resultant^ it bi- 
sects the angle BOB, R Fio. 16S. 

/. l^AOq, QOB, BOB ==2 rt. z-= z AOP, POS, SOB ; 

.-. iQOB^iPOS. 

The four forces are not necessarily equal, for the magnitude of the 
resultants of each pair, depending partly on the magnitude of the 
forces, and partly on the angle between the directions of the forces, 
may be equal, without equality existing between the four forces. 

2. P2 + Qa + 2Pg.cos^=iJ2=p2+g2 + 2Pg.cos(45"-^ ; 

/. cos^=cos(45*-^); 
.'. cosd=cos 45' . cos^ + sin 45' • sin^ ; 

.*. cos^= -T^ . cos^ + -jT . sin^ ; 

. ". V2 . cos^ = cos^ + sin^ ; 
/. i/2=l+tan^, or, tan^=V2-l. 

3. From A in BO produced draw AB parallel to P, meeting 0(i 
mB. 

P 




^a 



Then if in magnitude, 

22, Q, P are in descending order ; 

AO^ OB, BA are in descending order ; 

I OB A, L BAOy L AOB are in descending order ; 

.*. L POQf L BOB, L BOQ are in aaceindii\\^ ^TdkS^. 



62 



KEY TO ELEMENTARY STATICS. 



4. ML, NO, AC axe equivalent to MP, MB, ND, NP, AG 
which are equivalent to CN, ND, CM, MB, AC, 
which are equivalent to CD, CB, AC, 
which are equivalent to CA, AC, 
which represent forces in equilibrium. 

D N C 





Fia 166. 



5. Let h be the neight of the steeper plane, V the base, a the 
length, Pi and P% the powers employed. 

Now since the power on the steeper 

plane is equal to the resistance on the 

other, the height and base of the second 

" are respectively equal to the base and 

height of the first. 

Then since three times resistance on 
steeper = resistance on the other. 

a a ' 

.-. 9{a^-h?)=-h\ or, Za^sJlO.h. 
Then Pi : W^h : a=3 : >/10 ; 

p, : W^>Ja^^n? : a=^ :3a=l : VlO. 

6. Let be the given point within or without the triangle, O the 
centre of gravity 

A A 



w 




C B 

Flo. i«r. 




KEY TO ELEMENTARY STATICS. 



63 



Then GO is the resultant of QB, BO, 
GO is the resultant of OC, GO, 
GO is the resultant of GA, AO ; 
.-. three times GO is equivalent to GA, GB, GO, AO, BO, CO. 
Now GA, GB are exactly counteracted by GO, for if we produce 
CG to D, the middle point of AB, 

GA, GB have a resultant = 2^2) =0^ ; 
.'. three times GO is equivalent to ^0, BO, CO. 



7. Let E and be the centres of gravity of AMN and MBCN, 
G tiie centre of gravity of ABC. 
Let^O=x, ^i>=^. 

Now area of MBCN =3 times area of AMN ; 
.-. OGxZ=BGxl, 

Now OG=x-^ 

3 

BG^AG-AR=%h-'i.^=h.; 

.-. 3»-2^= A and /. x=^—h. 
«i 




8. Let AC=3 feet, CB=4feet, AB^b feet. 
Then z JLC^ is a right angle. 

Draw DN parallel to AC. 
Let ti, ^ be the tensions of CB, CA. 
Then CN :ND :DC=:ti:t^:25. 
But, by similar triangles, 

CN:ND:DC=^AC:CB:BA 
=3:4;5; 
/. ei:<a: 25=3:4:5; 

.'. ^=15 lbs., and ta»20 lbs. 




«5 

Fio. 169. 



64 



KE Y TO ELEMENTAR Y ST A TICS, 



9. The direction of W^ the weight of the rod 
BG^ bisects the angle BAC, 

The tensions of AB^ AC are equal, let each 
be represented by t. 
Then W^=^t^ + t^ + 2t.t ,coaBAC 
=<2 + t2 + 2i^cos60" 
=«2 + t« + 2i«.{ 

.•. tension of each string sbs-t-=—--^ . 






R 



Pio. 171. 



10. Let be the point on which the forces P, Q, R acting are at 
rest 

Then POiJ = (ISO* - 30") = 150'. 
And B:Q= bulPOQ : ^inPOJS ; (Art. xxxix.) 

.•.V3:l=sinP0g:J; 

.-. 8inP0g= ^ ; 
'. P0g=60% or 120". 



■*^ 



Hence ^OU- (360^ - 150" - 60") = 160', . . 
orC01J=(360;-160"-120")-=90"; . . 
.-. P : Q- 8inl50" : sinlSO", and /. P=ft 
or P : §- Mn90" : sinlW, and .-. P=2Q. 



Fio. 1. 
Fio. 2. 



KEY TQ ELEMENTARY STATICS. 



65 



11. First, when P acts horizontally 

Next, when the direction of P makes an angle of 30** with the 
plane, resolying in the direction of the plane, there is equilibrium 
when 

P.cos30**=Jr.cos60°, 
P cos60-*_l .V3_ 1 
W^ cos30'"" 2 • 2 "y3' 
which is the relation between P and W established in the first case. 

P 
Ri R2 / 

B 




y] 



W 



Fig. 172. 



Again, if i^ and i^ be the pressures on the plane, 

B^^ sin90^ _ sin90° ^g 
P sinl50* sin30' 
J^ 8inl50^ _, . 
P " sinl50" ' 

12. P : y : P=sinl20' : sinl05" : sinl35' 

(Art. xxxix) 



P 



=sin 60** : sin 75^ : sin 45* 

_>/3.V3 + l .JL 

2 • 272 V2 

^V6 . V3 + 1 2_ 
2V2 • 2^2 • 2^/2 

=V6:V3 + 1 :2. 

E 



/05* 




¥\<a. \1^ 



\ 



66 



KEY TO ELEMENTARY STATICS, 




Fio. 174. 



13. Let ^0 be the boat, AB the mooring 
chain, G the stake by which the rope JC is 
B fastened. Let the current act in the direction 
DA. 

Now, suppose AB to be cut. Then the boat 
is acted on by two forces, one acting along -40, 
the other along ACy and the resultant of these 
forces acts towards the bank GB^ to which the 
boat will therefore swing 




Fig. 175. 



14. Let oAy oBy oC represent the 
forces. 

Join AB, BG, GA. 

Complete the parallelograms AoBF, 
BoEGy AoCD, 

E Then CJAoBF^ EJFoBE 

(Euclid, L xxxvi.) 

=OoBj&(7; 

.-. i^AoB= lBoG. 

Similarly, each of these triangles 
= A AoG, 




15. ABy BChave a resultant 
AG. 

.'. ABy BG, DG are equiva- 
lent to AGy DGy of which the 
resultant is 2EGy t.e. EGF. 



Fig. 17a 



KEY TO ELEMENTARY STATICS. 



67 



16. Since the figure is symmetrical, pressure on B 
is equal to pressure on 0. 

Let P be the pressure on each of these pegs. 

Then the tensions of jBD, GJD will each equal P, g 
and zPDO= 60°. 



.-. Tr2=p3 + p2 + 2P.P.cos60°«3P2; 
.-. P= 



-^, or,P^ T-' 



Also, pressure on JL is clearly equal to W. 

17. The three forces acting on the sphere are 
the tension of the string AO, the pressure of the 
wall acting along the radius PO, at right angles 
to ABf and the weight of the sphere acting 
vertically through 0, and the lines of action of 
these forces are parallel to the sides of the tri- 
angle AOB, 

,\R:T:JV=OB:OA:AB 




W 
Pio. 178. 

18. Let T be the tension of the string, a the inclination of each 



plane, B the pressure on each 
plane, acting as a normal to the 
sphere. 

Resolving vertically, for the 
equilibrium of the sphere, 
pr=2P.cosa. 
Resolving horizontally, for the 
equilibrium of either plane, 
T=P.smo; 

.•.T=tana.-2. = y.^. 




68 



KEY TO ELEMENTARY STATICS. 



1 9. Let ^ (7 be the post, B its middle 
point, N^ 

I DBC= 30', and . •. z BBC= 60^ ^^ 
The angle between the two sets of 
telegraph wires is 120**, and if iS be 
the resultant of the equal tensions of 
the wires, each = T, 

222^ ^12 + >zi2 + 22^ . coslSO" ; 
. 222=3T2 and/. i2=T. 
Let t be the tension of rope BI). 
Then, taking moments round (7, 

i2x^(7-«xJ5a.cos60°; 

AG 
.-. TxAC=t X -^ X J, and .-. t=4T. 




Fra. 180. 



\ 



20. Let ABCD be the quadrilateral, and BF=CE, DG^AE, 
Bisect BC in 0. Join OA, OG, OB, 
Let H and K be the centres of gravity of 

ABG and BBC, Join B^ cutting OG in i. 

Then '.• OH^lOA, and OK^iOB ; 

.'. fix is parallel to AB ; 

.', OL=iOG, and L is the centre of gravity 
of triangle EFG, Draw AP and DQ per- 
pendicular to CB, 
Then triangle^JBO: triangle DC© = ^P : DQ 

=AE:BE 

=BG:AG 

=KL:LH. 

Therefore the triangles ABC and BOB will C 
balance about Z, or the centre of gravity of 
the quadrilateral is L ; and this is also the centre of gravity of the 
triangle EFG, 

21. Let A, B, C be the three pegs. 

z 5^0=120*; 
.-. z ABC= L AGB^W. 
Then pressure onA^2W. cos60'=Tr, 
pressure on B^2W, cos30'= fT. ^3 
= pressure on C, 




Fio. 181. 



KEY TO ELEMENTARY STATICS. 



69 



The yertical pressure on ^= TF acting npwards. 
The yertical pressure on both £ and C 

= PF+ W , co860*=-— acting downwards. 

The reactions which support the weights will 
be equal and opposite, so that we shall haye, in 

order to support the wei^ts, —^ at both B and C' 

acting upwards, and W acting downwards at A ; 

or 3 TT upwards, and W downwards, or 2 fF acting upwards, which 

will just support the weights. 




22. Resolving along Oa; and Oy, of which Or coincides with BOE^ 
observing that co660'= J and sin60'*=-^ 

Z=6 + 6.J-l.i-2-3.J + 2 
=6+3-J-2-|+2=6 



F=6 ^ + 1 ^ 



3.^g-4.4^=0. 
2 2 




Henoe the resultant acts along OEy and its magnitude is 6 lbs. 



70 



KEY TO ELEMENTARY STATICS. 



23. Let M be the middle point of the rod, W the weight of the 
^ _ rod, T the tension of each of the strings 

f AD, BE. 

J Take moments about C, and 
'^ T,AC=T.BC+W,MC; 

_ or, T. {AM+ MC) = T. {BM- MC) + W,MC; 
B and /. since T . AM=T,BM 

T.MC=W.MC-T.MC; 



M 



V 

w 

Fig. 184. 



W 




24. Let ABC be any one of the triangles, and 
AC, CB represent forces tending from A, and AB 
a force tending towards B, it is plain that the re- 
sultant of AC, CB will be represented by AB, and 
.'. the resultant of AC, CB, AB will be represented 



Fio. 185. by 2AB, 



25. Since the angles AEB, ABB are right angles, a circle described 
on AB as diameter passes through E and D ; 

.-. I ABE= L ad;e, 

and since the angles AFC, ADC are right 
angles, a circle described on ^ C as diameter 
passes through F and D ; 

.-. z ACF=^ L ADF. 
Now z ABE^ L ACF, 

by similar As ABE, ACF; 
^ .-. I ADE= I ADF,fixA.\ J jD bisects z FDE, 

*. forces are equal 




Fio. 186. 



» 




Fio. 187. 



3 26. Let AB, J.C be the arms of the lever. 
Let the forces P, P act along PD. 
Then the pressure on A will be represented by AD, 

Now ^jD2=2Pi>2 ; 

.-. ^2)=V2.Pi>, 
or pressure on A^ ^2 » T*. 



ELEMENTARY HYDROSTATICS. 



KEY. 



Examples — I. (p. 8.) 

(1.) 1 J : 64 = 1 ton : weight supported ; 

.•. weight = — - — tons = 56f tons. 



(2.) IJ : 240=3 cwt. : weight supported ; 

/. weight =^^^:i^^-:i^ cwt. =600 cwt. = 30 tons. 



(3.) Area of small piston : area of large piston = (1})2 : (60)* 

=9x9:50x60x64 
.-. weight supported ^ 50x50x64x 15 ^^^^^ ^ 800000 ^^^ 

= 29629-S2&lbs. 

(4.) Let X represent the pressure in lbs. 

Then2:144=a;:4j; 

2x9 1 
/. x=---- — x=r7: , .'. pressure will be 1 ounce. 
144 X 2 16 



(6.) Let % represent the pressure in lbs. 

Thenl :144=a;:9 ; 

9 1 
.'. «=——=—, .•. pressure will be 1 ounce. 
144 16 



I 



72 KEY TO ELEMENTAR Y HYDROSTATICS. 



(6.) Area of horizontal section of large cylinder = = ' 

sqnare inches. 
Let X s measure of lifting power in cwts. 

mu 11 10x10x22 on 
Then If : ^=20 :x; 

•*• **^ — 5q — cwt. =s 6687 J J cwt. 



Examples— II. (p. i8.) 
(1.) Pressure at 42} feet ^%^ • pressure at 32 feet 

^^® X 15 lbs. =20 lbs. 



32x3 



(2.) Pressure at 20} feet ==^ • pressure at 8 feet 

^^ x^ lbs. =37 A lbs. 



^ 



8x2 3 

(3.) The pressure at a depth of 8 inches in the second fluid Is 
double the pressure at a depth of 4 inches, and is therefore equal to 
the pressure at a depth of 6 inches in the first fluid. 

.'. the pressures are as 7 : 6. 

(4) The pressure at a depth of 12 inches in* the second fluid is four 
times the pressure at a depth of 3 inches, and is therefore equal to 
the pressure at a depth of 8 inches in the first fluid. 

.'. the pressures are as 9 : 8. 

(5.) Let X be the height of the column in feet. Then (30 - x) feet 
is the depth of the top of the column, and 30 feet is the depth of the 
bottom of the column. 

.-. 30-x:30=2:3, 
or, 90 - 3x=60, and .'. «= 10 feet 

6.) Pressure at 12 feet — \\ pressure at 15 feet 

= if X 15 lbs. = 12 lbs. 

(7.) Pressure at 12 feet = ^ pressure at 4 feet 

= yx31bs.=91b«. 



KE Y TO ELEMENTAR Y HYDROSTA TICS. 73 




(8.) Cubic content of water =(20 x 2J) cubic feet = 50 cubic feet ; 
.'. pressure =(50 x 1000) ounces =3126 lbs. = 1 ton 7 cwt. 3 qrs. 17 lbs. 

(9.) Cubic content of mercury = (8 x 3) cubic inches ; 

13600x24 1700 n iv los 

.*. pressure = — :r=^ — ounces =—x— ounces=ll lbs. 12§ ounces. 

(10.) Cubic content of water = (24 x 18) cubic feet ; 

.-. pressure =(24 x 15 x 1000) ounces =22500 lbs. 

(11.) Let ABC represent the base of the cistern : AD a perpen- 
dicular on BC, C D B 

Thsm AD^BB . ^Z. 

Hence, if each side of the triangle be 6 feet, 
area of base=^D . BD^QaJS ; 

.'. cubic content of water = (9 ^3 x 2) cubic feet ; 

, 9V3x2xl000„ -,__ ,-„ 
.*. pressure on base=-^^^^ r^ lbs. = 1125^3 lbs. 

(12.) When the spout has been broken midway, its upper extre- 
mity is three-fourths of the height of the tea-pot, and therefore this 
is the height to which the tea-pot can be filled. 

(13.) Since the external pressure on the cork increases as the bottle 
sinks, while the internal pressure on the cork is constant, the cork 
will be forced in when the external pressure exceeds the internal 
pressure. 

(14.) Pressure on bottom 

= (112 X 4 X 1000) ounces= Y^^Y^^3 tons = 12 J tons. 

(15.) Let X be the height of the water in the tube in inches. 

The pressure produced by this on the base of the piston, which is a 
square inch in area, will be equal to the weight of x cubic inches of 
water, which has to lift 7 lbs. 13 ounces, or 125 ounces ; 

X X 1000 



1728 



= 125 



1728x125 «,^. , ,«- , 
3?= — j^TQTr — = 216 mches=18 feet. 



74 KE Y TO ELEMENTAR Y HYDROSTATICS. 



Examples — III. (p. 27.) 
(1.) 1 cubic foot of copper weighs 8*91 x 1000 ounces, or, 8910 ounces. 
1 cubic inch of copper weighs r^^^ ounces. 

512 cubic inches of copper weigh — Tna^ — ounces, or, 165 lbs. 

(2.) 1 cubic inch of iron weighs as much as 38 cubic inches of 
amber. 

.*. specific gravity of iron : specific gravity of amber = 18 : 1. 



(3.) 1 cubic foot of mercury weighs 13500 ounces. 

1350C 
1728 



1 cubic inch of mercury weighs ounces, or, 7} J ounces. 



> 



(4.) 1 cubic foot of the substance weighs 50 lbs. 
1 cubic foot of the substance weighs 800 ounces, 
and 1 cubic foot of water weighs 1000 ounces ; 
.'. specific gravity of the substance is '8. 

(5.) 1 cubic foot of cork weighs '24 x 1000 ounces =240 ounces. 

1 cubic inch of cork weighs r^^ ounces. 

200 X 36 
36 cubic inches of cork weigh —ry-— ounces, or, 5 ounces. 

(6.) A cubic foot of the substance weighs 3000 ounces. 

A cubic inch of the substance weighs yfoQ ounces, or, If .| ounces. 

(7.) 1 cubic foot of the body weighs — lbs. 

m 

1 cubic foot of the body weighs - ounces, 

and 1 cubic foot of water weighs 1000 ounces ; 

/. specific gravity of body : 1 = — ; 1000 ; 

m 

.-. specific gravity of body = -^^^ = '^^. 
^ ^ ^ ^ 1000m m 



KEY TO ELEMENTAR Y HYDRO STA TICS. 75 

(8.^ 1 cubic inch of iron weighs 4} ounces, 

1 cubic foot of iron weighs (4J x 1728) ounces, or, 7776 ounces ; 
.*. specific gravity of iron : 1 = 7776 : 1000 ; 
.*. specific gravity of iron = 7*776. 

ft'7ft V 1 ft 

(9.) 1 cubic foot of the wood weighs — — — ounces, or, 1166*6 

1^ 

ounces; 

.*. specific gravity of the wood : 1 = 1166 '6 : 1000 ; 
.'. specific gravity of wood = l"l5. 

274-*? X 1 fi 
(10.) 1 cubic foot of ash weighs — -- — -— ounces, or, 844 ounces. 

.'. specific gravity of ash : 1 =844 : 1000 ; 
.'. specific gravity of ash = '844. 

(11.) Let V be the volume of the metal, and b the specific gravity 
of the compound. 

Then -|- is the volume of the alloy ; 

.-. vxl5 + |-xl2=rv +-|-)^« 
.-. 15 + 6=^, and.*. «= 14. 

(12.) Let 8 be the specific gravity of the lump. 

As 
Then --- is the specific gravity of one metal (for 2 : lj=4 : 3), 

and -^ is the specific gravity of the other (for 2 : 2J^=4 : 5). 
Let Vi and v^ be the volumes. 

Then Vi x -^ + vg x — = (vi + r^) x « ; 
3 5 

.-. 20Vi + 12Va=15t;i + 16«a; 
.'. 6vi=3va, or, Vi : t7a=3 : 5. 

Hence to form 2 cubic inches of the compound we must put | cubic 
inches of first and { cubic inches of second. 



76 KEY TO ELEMEJSITAR Y HYDRO STA TICS. 

(13.) Let Vi and v^ be the yolumes, tOi and w^ the weights. 
Then VbVi + Zv^=\vi-\-Vt} 2-6, 

or, 15vi + 30t;j= 26t?i + 25^2 ; 

and /. Vi : «;j=6 : 10=1 : 2. 

^^^'''r6"*"¥""2^' 

lOtOi Wa IOWi+IOk^ 
'''"15"'*"3"= 25 ' 

or, 60wi + 25u?a = 30Wi + GOwj ; 

.*. 20wi=6tr2 ; 

.*. tPi : Wa=5 :20=1 :4. 

(14.) Let a; and y be the measures of sea-water and fresh water 
respectively. 

Then x x 1*027 + y x 1 == (sc + y) x 1*009, 

or, l*027a; + y=l*009a; + l*009y, 

or, -OlSx^'OOOy; 

.'. X :y=l :2; 

that is, proportion of fresh water to be added = 2:1. 

(15.) Let w be the measure of the weight of each substance, d the 
density of the compound. 

mi,^„ w , to 2m; 

„ 100^100_2 
'325'^276~d' 

(16.) Let V be the measure of the volume of each substance, s the 
specific gravity of the compound. 
Then v X 2*5 +t7 X l*5 = 2v X « ; 

.*. 2*5 + 1*5=2«, and .•. «=2. 

(17.) Let s be the specific gravity of the compound. 
Then 5 x 11*35 + 5 x 7*3= 10« ; 

or. 11*35 + 7-3=2«, and .*. 8=9*325. 



\ 



KEY TO ELEMENTARY HYDROSTATICS. 77 

(18.) Let V be the measnie of the volume of each fluid, (2i, 4i9 (2s the 
measures of their densities, d the measure of the density of the 
mixture. 

Then vii + t;(2s + 1;(^ ss (i; + 1; + 1;)({ ; 

.'. (ii + d8+(t=3(i, and.'. (i,=3<i-(ii-i». 

(19.) Let « be the specific gravity of the compound. 
Then 10 x 8-9 + 7 x 7-3=17«, 

or, 89 + 51-l = 17«; 
.-. 17«=140-1, and .'. «=8-241 . . . 

(20.) Let % be the specific gravity of the mixture. 
Then^ + A + l=^-±6±7 
•7 -8 -9 B ' 

^^ 60^15^70 18 

.•.?^«l?,and.-.«=:-802 . . . 
126 « 

(21.) Let % be the specific gravity of standard gold. 
Then 11 x 19*3 + 1 x 862= 12« ; 

.•.212-3 + 8-62=12* ; 
.-.«= 18-41. 

(22.) Let « be the specific gravity of the mixture. 
Then 63 x 1-82 + 24 x 1 =86«, 

or, 114*66 + 24=86* ; 
.-. «=1-61 ... 

(23.) Let 9 be the specific gravity of the compound. 
Then4x2 + 6x3 + 6x4=(4 + 5 + 6)«, 

or, 8+15 + 24=15*; 
/. *b3-1S. 

(24.) Let 10 be the weight of gold in ounces. 
Then _!g__^ ll'5-u7 ^ir5 
19-35^ 2-62 7-43' 

^, 100M7 1150-100117 1150 
' 1935 ^ 262 743 ' 

20W 575-50117 ^1150 
' 387 ■*■ 131 743 ' 



• • — 



262 0?g + 222525 - 193 50M7 1150 
50697 " 743' 



78 KE Y TO ELEMENTAR Y HYDROSTA TICS. 



or. 



222526 - 16730tr 1150 



' 50697 "" 743 ' 
or, 165336075 -12430390tr=58301550; 

.-. 107034525 =12430390u7, 

and /. t(;=8'6 . . . ounces. 

(25.) Let tTi, w^ be the weights of iron and gold, Vi, v% the volumes 
of iron and gold. 

Then 7-8t?i + 19-4*,«8 (vi + t;,), 

or, ll'4t?a=*2t7i; 

.*. t;i:t;j=ll'4:-2 
=57 : 1. 

Again, 15. + J£!-=^_±i^, 
^^ '7-8 19-4 8 ' 

'''^' 39"*'W"~"8~' 

, 485ti7i + 195wa _ tPt + tOg 
•'• 3783 8 ' 

or, 3880k?i + 1560i«a= 3783wi + 3783t«?» ; 
.-. 97«\=2223wa; 
•^ tri:w2=2223 :97. 



ExAMPLBS— IV. (p. 42). 

(1.) Weight of water displaced =^ of weight of glass ; 

.'. specific gravity of water=T^ of specific gravity of glass ; 
.*. specific gravity of glass = V* = 3'S. 

(2.) Pressure 

= weight of {(28 x 1760 x 3) x J x 480} cubic feet of sea-water. 

28 X 1760 X 3 X 480 X 1-026 x 1000 

ounces 



4 
7 X 1760 X 3 X 480 X 1026 



tons. 



16x112x20 
(110 X 9 X 613) tons=507870 tons. 



KEY TO ELEMENTAR V HyDROSTA TICS. 79 

(3.) Let a;=the part immersed, t7= volume of body, 
Thena;:v=3-3:4-4; 

/. x—-r-. v= 77 v=a -J- of whole body. 
4*4 44 4 ^ 

(4.) Let x be the number of grains a sovereign weighs in water. 

Then 122'5 - a: = weight of water displaced ; 

.M9-4 : 1 = 122-5 : 122-5 - ar, 

or, 2376-5 - 19-4a;= 122*5 ; 

2376-5-122-5 .,-,« . . ._^ o^io 
• • *■* — "To^i =116Jf grams =4 dwt 20Jf grams. 

(5.) Weight of water displaced =2 lbs. ; 

.'. specific gravity of substance : 1 =8 : 2 ; 
.'. specific gravity of substance =4. 

(6.) Let X be the weight required, w the weight of the tub. 

Then since weight of water displaced by the whole tub=s4«;, 
a;=4w-tt;==3w— three times weight of the tub. 



(7.) Part immersed : whole body =1*4 : 2*1 ; 

14 

gT of whole body— ^ 



14 2 

.-.-part immersed=— of whole body =-5- of body. 



(8.) Weight of water displaced by leads=YT:^ ounces ; 
. •. pressure on bottom = ( 1 - yrri ) ounces = - - ounces = r= ounces. 

(9.) Half a cubic inch of water weighs y=^ ounces, 

a cubic inch of cork weighs 3-;=^ ounces ; 

17^0 

• 1.x X V jj J 500-240 65 

.*. weight to be addea= — - ounces=— ounces. 

(10.) Since the specific gravity of the cork is one-fourth of the 
specific gravity of water, weight of water displaced by the cork = four 
times weight of the cork. 

.'. tension of strings three times weight of the corks 3 ounces. 



8o KE Y TO ELEMENTAR Y HYDRO STA TICS. 



46 
(11.) Weight of water displaced by lead—rrig ounces ; 

.-. tension of string= ^46 - —^ ounces* ^46 x ^l"^ ~ ^ \ 



ounces 

0/ \ ll'O / 

46 X 105 



115 



ounces » 42 ounces. 



(1000\ 
•24 X Ytoq) o^^ces=^^^ ounces ; 



240 
1728 

^ , , . , , . 240 5 

.'. we must add a weight of p^Ho ounces, or, 5^ ounces. 



(13.) Weight of water displaced ==| of half the weight of the cube 

ss j of weight of the cube ; 
.% tension of string ={ weight of cube={ lbs. 

(14.) Weight of water displaced = (42 - 30) ounces = 12 ounces ; 
.'. specific gravity of substance : 1&=42 : 12 ; 

42 
.*. specific gravity of substance «=—= 3*5. 

(15.) 14 lbs. « (14 X 16) ounces=224 ounces. 
.'. weight of water displaced =(2560 -224) ounces =2336 ounces ; 
.'. specific gravity of substance : ls=2560 : 2336 ; 

.'. specific gravity of substance=-— 5-=— -. 

2336 73 

(16.) Weight of water displaced by substance 

= (20 + 12 - 18) ounces= 14 ounces ; 

.'. specific gravity of substance : 1 = 12 : 14 ; 

12 6 
.'. specific gravity of substanceos--— __. 

14 7 

(17.) Weight of water displaced by mahogany 

« (375 + 380 - 300) grains=455 grains ; 
.*. specific gravity of mahogany : 1 = 375 : 455 ; 

.'. specific gravity of mahoganyso -^a:-- . 



KE V TO ELEMENTAR V HYDRO STA TICS. 8 1 

(18.) Weight of water displaced by metal 

=(120-113) grains = 7 grains; 

.'. specific gravity of metal : 1 = 120 : 7 ; 

120 
.'. specific gravity of metal = ,,- ■" 17 J. 

(19.) Weight of water displaced by spar 

= (190 - 120) grain8=70 grains ; 
.•. specific gravity of spar : 1 = 190 : 70 ; 

/. specific gravity of spar = . = — =r2f • 

(20.) Weight of water displaced by former body 

= (4 + 3 — 2^) ounces = 4| ounces ; 
/. specific gravity of former body : 1 =4 : 4| ; 

A. 1ft 

/. specific gravity of former body=-^=-- . 

(21.) Weight of water displaced by wood and lead 

= (22+12-8) lbs. = 26 lbs. 
Weight of water displaced by lead 

22 „ 2200 „ 440 „ 
"n-35 ^^«- = 1135 ^^'- =227 ^^'•' 

.•, weight of water displaced by wood= (26 - o3y ) l^s* = 0017 ^^^ 5 

5462 
.*. specific gravity of wood : 1 = 12 : gg^ J 

.. ', f A 12 ^ 227 1362 

/. specific gravity of wood= ^^^ ^2731 * 

(22.) Let tc= weight of substance in vacuo, 2ia= weight of sinker 

in vacuo, «= specific gravity of substance ; 

.'. {w + 2w) — 2wy or, w= weight lost by both in water. 

Let t;= volume of sinker, then v x 1 = weight of water displaced by 

sinker; 

/. IP- v= weight lost by substance ; 

. w-v 1 
• • ^ — J 

W 8 

VS — V 1 , 8—11 tx 

or, — , and .'. ^— ^=— , or «-2. 

V8 8' 8 8 



i 



82 KE Y TO ELEMENTAR Y HYDRO ST A TICS, 

(23.) A cubic foot of cork weighs ('24 x 1000) ounces, or 240 
ounces ; .'.to displace a cubic foot of water, weighing 1000 ounces, 
we must increase the weight of the cork by 760 ounces, or 47} lbs. 

(24.) Let x^Yolume of part immersed. 
Thenaj:a: + 8=1 :3; 

Hence whole length of the cylinder = 12 feet, 

(25.) Four-fifths of the cylinder are immersed. 
Let 8 be the specific gravity of the cylinder. 

Then« : '825=— : 1 ; 

5 

.•..=il^=4x-165 = m 
5 

(26.) Because the specific gravity of salt water is greater than that 
of fresh water. 

(27.) Let X parts of an inch be in the mercury, then \-x parts of 
an inch are in the water. 

Then ac X 13-6 + (1 - a) x 1 = 1 x 7-8, 

orl3-6a;+l-a;=7'8; 
/. 12-6a;=6-8; 

• fl. 5§.-_3^ inches 

..a-— --mcnes. 

(28.) Since the specific gravity of the body is half the specific 
gravity of water, it will float with 500 ounces of its weight above the 
surface. 
And since 1 cubic foot of water weighs 1000 ounces, 

1 cubic foot of the body weighs 500 ounces ; 
.'. 1728 cubic inches of the body are above the surface. 

(29.) 12cubic inches of water weigh — ^ - — ounces, or, — ounces, 

1728 18 

or, 6J5 ounces ; 

.'. weight of body in water = 8 lbs. -6J| ounces, or, 7 lbs, ^^ 
ounces. 



KEY TO ELEMENTAR V HYDRO ST A TICS. ^3 



(30.) Let X =■ edge of the cube in inches ; 

/. 1 X X* = cubic inches of water displaced by the cube sinking 

an inch; 

1000 
• • ^ ^ V70Q~^®^^^ ^^ ^^ water in ounces ; 



1728 



/. x^x 



1000^1000 
1728 3 



.*. x2=576, and «=24 inches =2 feet. 

(31.) Content of box externally = (10 x 8 x 6) cubic inches = 480 
cubic inches. 

Content of oox internally = (9^x7^x5^) cubic inches rs'ii^ 

8 

cubic inches. 
.'. content of substance =f 



480 Q— ) cubic inches = -^ cubic 



inches. 

Hence, if « be the specific gravity of substance, 

705« .Q^ , 480x8 3840 

— =480x1, or,. =-^^=-^; 

/. «=5ff. 

(32.) Let DE be the surface of the fluid. 
Th&nADEiAFG^l:^, 

(Hydrostatics, Art. 63.) ^ 
But ADE:AFG^A&:AC^. (Euclid, VL xix.) 
.-.^52:^02=1:3; 

(33.) BE being the surface of the fluid, and J>, ^the middle points 
of AB, AC, triangle ADE = one-fourth of triangle ABC. 

A B C 





Fig. ISO. 




B C 

Now, part immersed : whole = specific gravity of body : specific 
gravity of fluid. 



84 KE Y TO ELEMENTAR V HYDROSTATICS. 



/. when the vertex is downwards, specific gravity of lamina = one- 
fourth specific gravity of fluid ; and when the vertex is upwards, 
specific gravity of lamina = three-fourths specific gravity of fluid. 

(34.) Let X be the number of pounds in the weight, « the specific 
gravity of the fluid, v and -^ the volumes of fluid displaced. 

Then 8 + ic=v.«, 
and 8=-5-.«; 

...®±5=3, ora;=161bs. 
(35.) Let « be the specific gravity of the cylinder. 



Then -I : 1=8:1; 



.-. 5=1 = -75. 



(36.) Volume of cube = (^A x A x -I'i cubic feet = ^ cubic feet; 

27 
,\ _J. cubic feet of the material weigh 2250 ounces ; 

o 

.*. 1 cubic foot of the material weighs — - ounces • 

o 

2 
.*. specific gravity of the material is — • 

Then if a; be the distance in inches to which the cylinder sinks, 

a;:3 = 4:l» 
3 

and .'. x=2 inches. 

(37.) Weight of fluid displaced = 1 lb. 

Then, if z be the specific gravity of the fluid, 

1 :3=«:2'7; 
.% 3«=2*7, or«=-9. 

(38.) Let v be the volume of the part immersed in cubic inches. 
Then v; 1=2-6: 13-5; 

.-..= 21^26- 
13-5 135 

26 
.*. the aluminium will sink to a depth of -—- incL 

135 



KE Y TO ELEMENTAR V HYDRO ST A TICS, 85 



(39.) Let w be the weight of the body in pounds. 
Then«;:t/; + 2 = — :1 ; 

.-. i£?-±_7J», or, 117=4 lbs. 

(40.) Let p be the pressure in pounds. 
Then 3:3+2?=^ :1, 

or3=^?, or,|)=31b8. 

(41.) 2 cubic feet of water weigh 2000 ounces. 

2 cubic feet of cork weigh 480 ounces. 
.*. tension of string =(2000 - 480) ounces =1520 ounces =95 lbs. 

(42.) Let «i and «, be the specific gravity of the bodies, s the specific 
gravity of the fluid. 

Then _!^-=.?L^ 
tPi — ti; s 

(43.) If the specific gravity of the wood be less than the specific 
gravity of the water, the water filling up the cavity will cause an 
increase in the apparent weight of the bullet. , 

(44.) Weight of first fluid displaced =3 lbs., weight of second fluid 
displaced =2 lbs. 

If then 81 and s^ be the specific gravity of the two fluids, and s the 
specific gravity of the body, 

-^=— ,and^ = -; 

.'. dividing, -i== — 

(45.) Weight of fluid displaced by the body = 1 lb. ; 
.'. if « be the specific gravity of the fluid, 

« 1 A . 7-7 , , 
777=y>and--»=y=l*l. 



i 



86 KE Y TO ELEMENTAR Y HYDRO STA TICS. 

(46.) Let 8i be the specific gravity of the first sphere, and 8^ be the 
specific gravity of the second sphere. 

Then 4«i is the specific gravity of the fluid. 
Let 2i; be the volume of the first sphere. 

Then Zv is the volume of the second sphere, 

and 2i; . «i + 3i; . «8=(2t; + 3i;) . 4«i ; 

.'. 2«i + 3s3=20«i; 

^47.) Let w be the weight in grains of the copper in air. 

Then ^--§?7^ 1 . 
w 8-85 ' 

.-. 885w - 887 X 885 = lOOw; ; 

.•.785^=887x885; 

.-.^=1-56999. 

157 

Then if 8 be the specific gravity of the alcohol, 

. : 1 = ^^^ _ 910 : 15^ _ 887 (Hydrostatics, Art 68.) 
157 157 

... ,=y69_99.ziM870^14129^.g ^^^ 

156999-139259 17740 ^ 

(43.) Let X be the depth in inches to which it sinks in alcohoL 
Then a; X -79=4x1 ; 

•*• ^''•79''y9 " ^^ ^^^^ 

(49.) Five lbs. of the compound are immersed. Let x be the 
number of lbs. of silver in the part immersed, b-z the number of 
lbs. of aluminium in the part immersed. 

Then since 10 lbs. of mercury are displaced 

X b-x 10 
+ 



10-4 2-6 13-5' 

X b-x _ 10 

^^'104 ~26"""135' 

. 20-3a; ^2 , 

* * 104 27 ' 

.'. 640- 81x=208, or, a;=4^ lbs. ; 

.*. there are 82? lbs. of silver in the whole mass. 



KE Y TO ELEMENTAR V HYDRO STA TICS. 87 

(50.) Let t«7= weight in pounds of water displaced by the rod. 
Then a; : 5 = 1 : 7-8 ; 

_50_25 
••^ 78""39* 

Hence tension of string = no - 55 ) lbs. = 9Jf lbs. 

(51.) Weight of water displaced =(1 - "905) ounces = '095 ounces ; 
.*. if « be the specific gravity of the silver, 

J. = _l =1222. 
1 -095 95 ' 

.-. «=iofy. 

(52.) Weight of water displaced by first body = J lb., and, if « be 

the specific gravity of the second body, weight of water displaced by 

2 
second body = — ^Ibs. ; 

.*. tension of first string = J lb., 

and tension of second strings 1 2 ) lbs. 

Now these tensions must be equal ; 

.-. -5-=2 ,and.\ «=»-^=l-8. 

(53.) Let X inches be above the surface. 
Then9-a;:9=i- :1; 

.*. 9 -a;=3, and .'. a;=6 inches. 

(54.) If 8 and ^ be the specific gravity of the fluids, V and F' the 
parts immersed. 

5 : iSr= F' : F. (Hydrostatics, Art 66.) 

=m-y :m-x. 

(55.) The weight of fluid displaced by body and lead =7 lbs. 
Now 5 J lbs. of lead displace J lb. of water ; 

.'. 5} lbs. of lead displace 2 lbs. of the fluid ; 
.*. weight of fluid displaced by body =5 lbs. 
Then if « be the specific gravity of the body, 

«:4=3:5;* 
12_ 

D 



I 



88 KEY TO ELEMENTARY HYDROSTATICS. 

(56.) Let i^= weight of the substance in air in ounces. 

Th ^~^^ _ sp ecific gravi t y of wate r _ 1 
w- 15 "specific gravity of alcohol" *7947' 

(Hydrostatics, Art 68.) 

.-. •7947w-7*947=i^- 15 ; 
.-. 7-053= -205311?, or, v)^^^^ ounces. 

Hence, weight of water displaced by substance 

/70530 -A 

^^Uosa-^V^^"^ 

50000 
= 2053^^^^^' 

•*. if X be the number of cubic inches in the substanoCi 

lOOOg^SOOOO 
1728 2053 ' 

50 X 1728 86400 , . . , 
••• ^=^053- = 2063 "^^^" "^"^^- 

(57.) Let X cubic yards be the volume of the granite, (1 — x) cubic 
yards is the volume of the ice. 

Then x x 2*65 + (1 - x) x '918=11 x 1 ; 

or, 2-65x + -918 - •918x= '92 ; 
.*. l-732x=*002; 

•*• *= j^3 cubic yard. 
866 

(58.) 320 ounces =20 lbs. ; 

.*. specific gravity of the wood : specific gravity of water =20 : 12. 
Hence if x be the part of the wood below the surface, 

x:l = 12 :20; 

••'^-20- 5 

(59.) 3 inches of the wood weigh as much as 2| inches of water ; 
.'. 1 inch of the wood weighs as much as | inch of water ; 

,•, 1728 inches of the wood weigh -^r — ounces, or, 750 ounces. 



KE V TO ELEMENTAR Y HYDROSTA TICS. 89 

(60.) Let X parts of a cubic foot be immersed. 
Then x: 1=85 : 103; 

. „^ '85 ^ 85 
1-03 103' 

(61.) Let X be the number of feet in an edge of the cube. 
Then 7? (x- 100) : x3=-9214 : 1*0263 ; 

/. 10263a;- 102-63= •9214X; 
.•.•1049x= 102-63; 

^. ^^1026300^^^3.3 ^^^ 
1049 

/. x3=936302451-687 nearly. 

(62.) Let X represent the pressure. 
Then t/; + x : ie7=4 : 3 ; 

.-. 3M; + 3x=4ti7, and .*. x=-—» 

(63.) The Yolume of fluid, v, displaced is the same in each case. 
Hence, if s and tf be the specific gravity of the fluids respectively 

V8 :v8'=rri in ; 
and .*. 8 \8'—min, 

(64) Let V be the volume in inches of the part immersed in water. 
Theni; :i; + 4 = l : 3; 

.*. 3i?=i? + 4, ori;=2. 
Hence, if x be the weight of the hydrometer in grains^ 

a: + 40 :x=2 :2-:^; 

or (a: + 40)x23=24x; 
.*. x= 920 grains. 

(65.) 980 divisions are below the surface in distilled water^ 954 
divisions are below the surface in sea water ; 

.*. if « be the specific gravity of sea water^ 
«: 1=980: 954; 

. 980_490_ 13 _,.^272 
••'"^954 477^^-^^^^^ 



90 KEY TO ELEMENTAR Y HYDROSTATICS. 

(66.) Weightof waterdisplacedby wood«(13 + 6-8) lbs. = ll lbs.; 
/. if « be the specific gravity of the wood, 
«:1=:6:11; 

.•.«=!= -54. 

(67.) Let X be the weight of the hydrometer in grains, 
m, « + 67 1 
^^^-ir==-866' 

.-. •866a; + 58-022«a;; 
.-. -1343;= 58022; 

68022 ... . 
.*. a;= "104- =433 grains. 

(68.) B weighs in water (11 - 10) ounces, or, 1 ounce ; 
B weighs in air 15 ounces ; 
.'. if t? be the Yolume and « the specific gravity of B^ 

t>(g~ -00 13)^15, 

t;(«-l) 1 ' 

.-. «-'0013 = 15«-15. 

.-. 14«= 14*9987; 

.-. «= 1*0713 nearly. 

(69.) Let t«7 a weight of the substance in air. 

The ^~^Q — specific gravity of water _ 1 ^ 
t/; — 25 ""specific gravity of alcohol"" '7947 ' 

(Hydrostatics, Art 68.) 
.*. •7947w-15*894=w-25; 

.*. 9106= -205311;, and .*. '^^-i^^H - 2o) ounces. 

Hence weight of water displaced by substance 

/91060 ^^\ 

== I — 20 1 ounces 

V 2053 / ""^"^ 

50000 

= -TT ounces ; 

2053 ' 

•*. if X be the number of cubic inches in the substanoe 

1000a ; ^50000 

1728 2053 ' 

/. 35=-oQg«- cubic inches =422\f5^ cubic inches. 



KE Y TO ELEMENTAR V HYDROSTA TICS. 91 



Examples — V. (p. 61.) 

(1.) Let the measure of the capacity of the barrel be oe. 

Then the measure of the capacity of the receiver is lOx ; 

(lOx \® 
— I . original density 

»0. original density. 

(2.) As in Example (1.) 

density after two strokes = ( ) . original density 

VoX t" X/ 

= — times original density. 

(3.) Let X and y be the capacities of the receiver and the barrel 
• ^, / 35 V 729 

. _x 9_ 

•'•x + y~10' 

.*. 10a:=9x + 9y, and .'. x ; y=9 : L 

(4) No : because the pressure varies with the depth alone ; so that 
if the section varied there would still be equal vertical increments of 
space for equal increments of pressure. 

(5.) Pressure on a square inch = weight of a column of water whose 
height is 90 feet and base a square inch + weight of column of mercury 
whose height is 30 inches and base a square inch. 

= weight of (90 X Yj^ cubic feet of water + weight of 30 cubic 

inches of mercury 

90 X 1000 , 30 X 47 ,._^ , „... ^^^.. 

= — :rjT — ounces + — g — ounces = (625 + 235) ounces = 53| lbs. 



92 KE Y TO ELEMENTAR V HYDROSTA TICS, 



I 



P-- 




(6.) When the mercury stands at Pin the tube, 
let be the point at which it stands in the basin. 

Now let the mercury fall to 22 in the tube, and 
rise to i> in the basin. 

Then CD-=^PE; 
, Q .*. real fall = apparent fall + ^ of apparent fall 

= IJ inches + — of 1 J inches 
2 10 ** 



Fio. 191. 



= ( 2" "^ 2o) ^^^^ '^ ^^ inches. 



(7.) The mercury would fall to the level of the mercury in the 
basin if the hole were made above the column of mercury in the tube. 
If it were made in the part occupied by the mercury, it will drive 
part of the column up the tube, and the other part down. 

(8.) Pressure on of a square inch = 1 ounce ; 

.'. pressure on a square inch = 234 ounces = 14*625 lbs. 

(9.) Let X be the height of the column in inches. 
Then x x 3-4=30 x 13*6 ; 

30 X 136 
34 



x> 



oyj_2 ^OKJ ^ j2^ ^^^^ ^ ^^ ^^^ 



(10.) It will have no effect, because a volume of mercury equal to 
that displaced by the iron will descend and allow the iron to take its 
place without disturbing the general upper surface. 

(11.) The body floats in two fluids, air and water, and the absolute 
weight of the floating body is equal to the weight of the displaced 
fluids ; hence if the air be removed, more water must be displaced to 
support the body, and therefore the body sinks. 

(12.) If the air be admitted above the column of mercury it will 
drive it down a little, the extent of the fall depending on the quantity 
of air admitted. 



KEY TO ELEMENTAR Y HYDROSTATICS* 93 

(13.) A column of water 20 feet high represents a pressure of 

— of 15 lbs. on the square inch. 
34 

.*. if z be the area of the valve in square inches, 

20 X 15 X a; ,^ , ^, „ 
«7 lbs. = pressure on valve = 21 lbs. ; 

34x21 714 ^_ 
*** *~20x~i5~300~ square inches. 

(14.) 500 fathoms = (500 x 6 x 12) inches = 36000 inches ; 
.*. if (2i be the density of atmospheric air and d^ be the density of 
compressed air, 

<t : di = (30 X 13-57) : (36000 x 1 -027) + (30 x 13-57) 
= 407-1: 37379-1; 

•••*-3ySi='^^^^^^^^y- 

(15.) The water barometer under the same additional pressure 
would rise (20 x 12) inches, or 240 inches ; 

.-. the mercurial barometer will rise ,-;—-; inches, or inches. 

13'57 ' 1357 ' 

or, 17j%Yr iiiches, or 1 foot 5j^^V inches. 

(16.) When the floating body is partially immersed, both air and 
water are displaced ; but the absolute weight of floating body = weight 
of displaced fluids, which must therefore be constant. Therefore when 
the barometer rises, there must be a smaller water displacement^ that 
is, the body rises ; while a decrease in the atmospheric pressure, when 
the barometer falls, will necessitate an increased water displacement, 
and therefore the body will sink a little. 

(17.) Pressure at sur£eu»=the atmospheric pressure. 

Pressure at 15 inches below the sarface of mercury » 
(atmospheric pressure at surface of water) 
+ (pressure at depth of 17 feet of water) 
+ (pressure at depth of 15 inches of mercury) 
BB(atmo6i^eric pressure) +}> (atmospheric pressure) 

+ 1 (atmospheric pressure) 
■stwioe atmospheric pressure ; 
.'. the pressures are as 1 : 2. 



94 KE Y TO ELEMENTAR V HYDRO STA TICS. 

(18.) A balloon is a hollow envelope, niade of some light material, 
as silk, which, when filled with heated air, or some gas, rises in the 
air. This must be the result, if the weight of the air displaced by the 
balloon exceeds the weight of the balloon. 

(19.) The external pressure on the bladder, increasing as the bladder 
is forced lower into the water, compresses the bladder, till at last the 
volume occupied by the air in the bladder is less than a volume of 
water equal to the weight of the bladder and air contained therein, 
and then the bladder must sink. 

(20.) Let X be the height of the column in inches. 

Then pressure on a square inch 

= weight of X cubic inches of mercury 
= weight of 13*56ic cubic inches of water 

13-56a; x 1000 



1728 

13-56arxl000 
1728 



ounces. 



= 226; 



226 X 144 32544 „^ ^ . , 
••^=^1130 ^-ll3()-=28-^"^^^^- 



• • 



(21.) Area of piston =36 square inches. 
Original pressure on piston =(36 x 15) lbs. = 540 lbs. 
Then since the pressure of air varies inversely as the space it 
occupies, 

new upward pressure on piston : 540= 216 cubic inches : 72 cubic inches 

= 3:1; 
.*. upward pressure on piston =1620 lbs, 
and downward pressure on piston by atmosphere =540 lbs. ; 
.'. a weight of 1080 lbs. must be placed on it. 

(22.) Height of water barometer, when the pump ceases to work, 
is 364 inches ; 

.*. height of mercurial barometer =--— inches = «/» inches 

13*6 136 

B 26^ inches. 



KE V TO ELEMENTAR V HYDRO STA TICS. 95 

(23.) Height of column of mercury within the bottle = twice the 
height of mercurial column exposed to the ordinary pressure of the 
atmosphere = 60 inches =5 feet. 

(24.) Let X and y be the capacities of the receiver and the barrel. 

Then^^±^=|-; 
X 2 

.*. 2x + 6y — 3x, and .*. x :y = 6 : 1. 

(25.) The air will be compressed inside, and so will displace less 
water ; and since the cylinder floated before the increase of atmos- 
pheric pressure, it will now sink, because the weight of fluid displaced 
is now less than the weight of the cylinder. 

(26.) Let X and y be the capacities of the receiver and the cylinder. 
Pressure after twenty strokes =^-i—^- (original pressure) 

X 

by 
= five times original pressure. 

(27.) Let X be the depth in feet 
Then 31 :30 + a;=l :2, 

or, 62 = 30 + a;, and .'. 05=32 feet. 

(28.) Area of base of cube = 100 square inches, 

and pressure on base = n 000 x jwao) o^Jices ; 

. . pressure on a square inch of ba3e= (lO x ]^^ ounces 

= 5j®o\ ounces. 

(29.) Pressure on 16 square inches of surface = (236 x 16) ounces 
= 3776 ounces. 
Weight of column of water of (16 x 9 x 12) cubic inches 

= ( 16 X 9 X 12 X _- j ounces= 1000 ounces ; 

.*. whole pressure =4776 ounces. 



96 KEY TO ELEMENTAR V HYDROSTATICS. 

(30.) Let D be a point in the surface of the fluid, and DA 
vertical line. 

n „ T^rAoan ra of 7? DR 



a 



— c 

-B 

FiQ. 192. 



rp, pressure at .B _ DB ^ 
pressure at A DA ' 
. pressure at J. — pressu re at B__DA — DB 
pressure at -4 E2 

Sunilarly, 



_AB 
DA 



Sunilarly, 
pressure at ^ - press ure at G __ DA - DC A G 
pressure at A DA 

By division, 



DA' 



ivision, 

pressure at J. - p re ssure at B _ AB ^ 
pressure at JL — pressure at C7 AC* 
p AB 



9. 



B 



P 

Pro. 193. 



AC 

p AB AB 

"q-^p^AC-AB BC 

(31.) An outer barrel ACDB encloses an inner barrel 
OR into which the bullet is rammed. 

Air is injected through the end of the outer barrel 
through the stock P. When a quantity of air has been 
confined in the barrel ACDB, a trigger at JK" is pulled 
which opens the bottom of the inner barrel, and the air 
being suddenly admitted projects the bullet from B with 

^ much force. 

r\ 
Q 
(32.) iSfuppose the column of mercury 

to rise from P to Q in the tube. 

Then it sinks in the cup to 0, a point 

such that 00=-! of OP. 

17 ^ 



Hence the true variation is 

Pg + lofPftor,J|ofPg. 

Hence in graduating the scale the distances 
actually measured from the zero point must be less 
than the space indicated by the numbers placed 
a^inst the graduations in the ratio of 17 : 18. 



\U 



C 




Fxo. 104. 



KE V TO ELEMENTAR V HYDROS TA TICS: 97 

(33.) Pressure of air on a square inch 

= weight of a column of mercury resting on a square-inch base 

= weight of 30*5 cubic inches of mercury 

=weight of (30*5 x 13*57) cubic inches of water 

= 30*5 X 13*57 X 252*6 grains 

= 104547*351 grains. 
Now 7000 grains troy = a pound avoirdupois, and .*. 104547*351 
grains =14*935 lbs. nearly. 



(34.) The pressure of air in the upper part of the tube, when the 
instrument stands at 27 inches, is equal to the weight of 3a cubic 
inches of mercury, a being the area of the base of the column in the 
tube. 

The air occupies a space of 9a cubic inches. 

Hence if this air be subjected to a pressure of 30a cubic inches of 
mercury it will be reduced to atmospheric density, and if a; be the 
space it then occupies in cubic inches, 

xi9a=2a : 30a ; (Hydrostatics, Art. 80.) 

••^"30 ""10' 
.*, the air will occupy ^q of an inch in length. 



(35.) Actual weight of stones + weight of air displaced by stones 
= actual weight of the weights + weight of air displaced by weights. 

Now the weights are smaller in volume than the stones. 

.*. weight of air displaced by stones is greater than weight of air 
displaced by weights. 

And a diminution of atmospheric pressure will decrease the weight 
of air displaced in each case in proportion to the weight of air 'displaced 
before in each case. 

.*. it will decrease the weight of air displaced by stones more than 
it decreases the weight of air displaced by weights. 

.'. the stones will appear to weigh more than their actual weight ; 
.'. he will gain by selling when the barometer is low. 

O 



98 KEY TO ELEMENTARY HYDROSTATICS. 



(36.) The confined air is pressed by a column of mercury equal to \ 
atmospheric pressure^ and by the atmospheric pressure at the sur&ce 
of this mercury. 

.'. we have a body of air at atmospheric density subjected to a 
pressure | times the atmospheric pressure ; 

.*. the air will be compressed into § of the space it occupied before ; 
•*. the mercury descends } of 14 inches, or 4J inches. 



Examples — VI. (p. 75.) 

(1.) It will increase the time of fiUing the receiver, because the only 
effecti/oe work will be done by the descending piston after it has passed 
the hole. Hence, if the aperture be made one-third of the way up the 
barrel, the distance through which the piston acts effectively being 
only one-third of the distance through which it worked effectively at 
first, three times the original time will be required for filling the tank. 

(2.) The tension of the piston-rod = pressure of atmosphere above 
— pressure of atmosphere below. 

Now pressure of atmosphere above = (4 x 15) lbs. = 60 lbs., pressure 
of atmosphere below = (4 x 15) lbs. — weight of a column of water 16 
feet high resting on a base of 4 square inches 

= 60 lbs. - weight of n 6 x - 7, ) cubic feet of water ; 
.'. tension of rod = weight of — cubic feet of water. 

•7 

iaaa\ 4x1000 

1000) ounces = 



= (|xl000) 
=^|? lbs. =27 J lbs. 



9 X 16 lbs. 



(3.) (a) K the hole be made in the longer arm, hehw the level of 
the shorter arm, no effect will be produced. 

(P) If the hole be made in the longer arm, above the level of the 
shorter arm, all the fluid in the longer arm below the hole wiU 
descend, and all above in the same branch will ascend, causing the 
remainder of the fluid to flow through the short branch, till the siphon 
is emptied. 




KE V TO ELEMENTAR Y HYDR OSTA TICS. 99 

(y) If the hole be made in the shorter arm, all the fluid below the 
hole in this arm will descend ; all above in this arm will ascend and 
flow through thfe longer arm, emptying the siphon. 

(d) If the hole be made at the top of the siphon, the fluid will 
descend in each arm, and will empty the siphon. 

(4.) A height equal to the height of the water barometer at the time, 
which is (13-57 x 29) inches, or 393*53 inches, that is, 32 feet 9*63 
inches, or 32*79416 feet. 

(5.) The fluid would descend in each branch, and the siphon would 
be emptied. 

(6.) Equally well at both. 

(7.) No ; because the water in the hold must be below the level of 
the surface of the water in the harbour. 

(8.) The height of the water barometer at the time, and this is 
(taking the specific gravity of mercury as 13*57), (30 x 13'57) inches, 
or 407'1 inches, that is, 33 feet 11*1 inches. 

(9.) If the air be removed from the siphon, the fluids will first 
ascend in each arm, and then the flow from the longer arm will 
commence and go on in the usual manner. 

(10.) The water will rise in the inverted tube as high as the top of 
the inserted tube, and then it will flow out of this tube. 

(11.) The water would soon cease to flow, because there would be 
no atmospheric pressure to cause a continuous flow. But when the 
air is gradually re-admitted, the water will begin to rise in each 
branch, and ultimately the flow will go on regularly. 

(12.) (a) The water will flow into the lower vessel. 

(6) The water will descend in each arm of the siphon till it 

stands at a height above each surface equal to the height 

of the water barometer, 
(c) The water will flow into the lower vessel 

(13.) The shorter arm must be 2 feet in length, and the longer arm 
just more than 2 feet. 



I 



loo J^EY TO ELEMENTAR Y HYDR OSTA TICS. 



Examples— VII. (p. 8i.) 
(1 .) 0= I {F- 32), and U = i.(J' - 32) ; 

.•.(l)C=|-(30-32)=^=-lJ. 
JJ=-i(30-32)=-|. 

.(2)0=-|(45-32)=|-xl3=7f. 
ie=i.(45-32)=-|-xl3=5f 

(3) (7=|-(56-32)=-| X 24 = 13J. 
i{=i.(56-32) = -| X 24=10§. 

(4)C=|(0-32)=^=-17f 



9 ^" " ' 9 
|(0-32)=-4? 



B=4(0-32) ^=-14J. 



(5)C=|<-7-32) ^=-21§. 

JJ=-i(-7-32)=^=-17J. 

(6) C=|(-45-32) = ^= -42$. 
iJ=|(-45-32) = -|??=-34S. 



(2.)C=5j,andJ'=^+32; 



/. (1) C=^=f =6i. 

^^=^+-32 = 11^ + 32=431, 



• KEY TO ELEMENTARY HYDROSTATICS. loi 






(3)C=°J«xO. 
^^=0 + 32=32. 

(4)0=^=-22J. 

^=^15? + 32=-40j + 32=-8}. 



(5)C=5iii^ = 5x(-16) 80. 

^=— ^~^'^' + 32 = (9x -16) + 32=-112. 



(6)0=^^=150. 

J'= ?-^i^ + 32 = 270 + 32 = 302. 



(3) ^^=-1- 0+32, and 1?=^; 

O 5 

q y 1ft 

.-.(1) ^=--g-+32 = 28J + 32 = 60jt. 
P_ 4xl6 _64 



(2) ^=?4^ + 32=81 + 32=113, 

D 

T> 4x45 . _ „^ 
iJ=—_ --=4x9=36. 


(3)jp=^i^ + 32 = 198 + 32=230. 
jR=i^^=4x 22=88. 



102 KEY TO ELEMENTARY HYDROSTATICS. 



(4)J'=?^+32=32. 

o 

(6) j'=?iLtil) + 32= -27+32-6. 

28-121^= -12. 
o 

(4.) TeSy if the graduations are to be onifomi. 

(5.) 9C=5 (J'-32) ) . 
0+ ^^=60 J ' 

.-.90=5 (60-0-32), 

or, 90=300-50-160; 

/. 140=140, and /. 0=10, and ^=6a 

(6.) 90=5(^-32) ) 

.-.90=5(50-32); 
/. 160=160, and /. 0=10, and J?'=60. 

(7.) 90=5 (J*- 32)) 

.-.90=5(0-32); 
.-. 40= - 160, and .'. 0= -40, and jP= -40. 

(8.) (80- 20)% or 60" on the new scale = (80- 32)', or 48"-P; 
.-. each degree on the new scale =--^ J*, or — ^- 

(9.)C=-|(-P'-32) 



9 



^ 



.-. 0-1 (78-3!)-?-^-^-i!5|. 



KEY TO ELEMENTAR Y HYDROSTATICS. 103 



(10.) 90= 5 (J?*- 32)) . 
(7+J'=88 J ' 

.-.9(7=5(88-0-32); 
.-. 140=280, and .-. 0=20, and ^=6a 



(11.) 0-|-(J'-32); 



.•.0=|- (49-32); 
••^-"9 9" ^* 



(12.) The graduations would be inconyenientlj smalL 

(la) 90=5 (J*- 32)) 
J>'=30 5 ' 

.-.90=5(30-32); 
.-. 6O-160, and .'. 0=265, "^d J'=8a 

(14) 9O=5F-160) . 
170= 5 J* S ' 

.-. 90=170-160, and .•. 0=20, and J?'=6a 

(15.) 90= 51^-160) 

.-. 90= - 50- 160, and .'. 0- - llj, and J^=llf. 

(16.) Let X be the number of degrees on the latter. 
rp, 15-9 a;-12 

^^^^i5:To=x-ri4> 

.-. 6(a;-14)=5(a;-12); 
.-. 6a;-84=5x-60, /. «=24 

(17.) Let X be the number of degrees on the latter. 
16-8 x-11 



Then 



16-10 x + 14' 

.-. 8(x-14)=6(x-.ll); 
/. 8x-112=6x-66, and.-. «— 23. 



1 



1 04 KEY TO ELEMENTAR Y HYDRO ST A TICS. 



aa)ii=-|(J'-32) 



\ 



.•.922=4(22 + 47-32); 

.-. 522=60, and .-. 22=12, and J?'=59. 

Also, if (2 be the number of degrees in the difference, 

22=4 (J?*- 
22= 



^=|(^-32)) 
;=J'-47-d ) 



.-. 4P- 128=9 (J?'-47-d), or, 5^=295 + 9(2, or, J'=59 + 4-(i; 

5 

.'. F riaes ?? and .*. 22 rises ~ 



Examples — ^VIII. (p. 87.) 

1. Let X be the yolume of the part immersed, y the volume of the 

part out of the water. 

m, a: -925 

Then = , ^,^ ; 

x + y 1-026' 

.-. 1025a; = 925a5 + 925y ; 

.-. 100a;=925i/; 

and ,\ y\x=\QO: 925=4 : 37. 

2. Let X be the volume of the body, y the volume of the part out 
of the fluid in the first case, s the specific gravity of the body. 

Then — -^-tt, 
X -9' 

and^=.-T ; 
■ X 1*1 * 

. x-y_V\^ 

•• y --9 ' 

. * 1-11 1. ?? 
"y '"9'^^'y = 9- 

^^°^20'n' 
and.-.«=|^=-495. 



KE Y TO ELEMENTAR V HYDROSTA TICS. 105 



3. f=^+32; 

5 



.-. whenC=-40% 

J^= -72 + 32= -40; 

and when 0=350% 

^^=630 + 32=662. 



4. After three strokes 

(3 \' 
- — -J . original density ; 

27 
.*. pressure =-— of original pressure ; 

27 
.*. height of barometer =_- of 28 inche3=ll*8125 inches. 

u4 



5. Let Wi and w^ be the weights of the hydrometers, v the volume 
of fluid displaced in each case, Si and ^ the specific gravity of the 
fluids. 

Thenii=.^>=-^=1. 

6. Let V be the volume of the man, s his specific gravity ; 

/. :^ is the volume of his head. 

\^=volu.eofmanunMe«ed. 

— = volume of bladders : 
12 

.*. volume of water displaced =——. 
Then v . « = -^^ . 1, or, « = {|= I'OSS. 

7. Height=(28 x 13*6) inches=380-8 inches=31 feet 8-8 inches. 

8. C-f- (i?'- 32) =1 (27-32)= -^=-2|. 



10(5 KEY TO ELEMENTARY HYDROSTATICS. 



I 



9. Let X be the depth in feet reached by his fingers, 

15 . 
jc - — is the depth reached by his feet ; 

15 
•*. a; : a;- — =3 :2^ 

or, 2a;=3a5- 22J, and .*. jc=22j feet 

10. Let X be the size of the cork in cubic feet. 

Then weight? of cork=240a5 ounces, and weight of mercury dis- 
placed =13600x ounces ; 

.-.« (13600 -240) = 167; 
^ 167 ^ 1 
•'•^ 13360 80* 

11. Let X be the number of degrees on De Lisle. 

150-x 47-32 



Then 



or, 



150 180 ' 

150 -a 15 



> R « > 



5 6 

.•. 900 - 6j5=75, or, 6x=825 ; 
.-.a; =137 J. 

12. Let «= volume of cork. 

Then since ('0013 x n)= specific gravity of n atmospheres, 

vx-24=wt;x-0013; 
•24 2400 



.*. n= 



•0013 13 



13. The body will rest when reduced to ^Z/ its natural size ; 

20 + n JL^ 

**• 20 + 471*" 2 ' 
.'. 40+2n=A20 + 4n; 

.'. n=10 feet, 

14 (a) When it is at depth of 5 feet, 

20 + n _ 20 + 5 _ 6 , 
20 + 4n~20 + 20~ 8 ' 

.•. body displaces — of 20 lbs. of water, or, 12 J lbs. of water ; 

8 

.•. since the body weighs 10 lbs., we must have a downward force 

of 2 J lbs. 



KEY TO ELEMENTARY HYDROSTATICS. 107 



03) When it is at depth of 30 feet, 

20 + n _ 20 + 30 _ 50 _^ ^ 
20 + 4n""20 + 120""l40""l4' 

.•. it displaces — of 20 lbs. of water, or, 1\ lbs. of water, 
.*. we must hare an upward force of 2f lbs. 



16. When the body displaces 11 lbs. and 9 lbs. of water respectively. 

20 + n 11 



(1) When 



(2) When 



20 + 4n 20' 

or, 400 + 20n « 220 + 44n, 
or, 24n=180, and .*. n=7J feet. 
20 + w 9 



20 + An 20' 

or, 400 + 20w = 180 + 36n, 
or, 16n=220, and .-. n=13| feet 



16. Let F be the reading on the Fahrenheit scale. 

Then -^ (^-32) is the reading on Reaumur's scale. 

•7 

.-. JP:i- (^-32)=25:4; 

.-.36^=100^-3200. 
Hence ^^=50. 



17. Let F be the reading on the Fahrenheit scale. 

Then -^ (-^- 32) is the reading on Reaumur's scale, 

a 

and -^ (#-32) is the reading on the Centigrade scale. 

.-. i^+ -| (J* - 32) + 1. (-F- 32) = 212 ; 

.% #+F-32=212; 
.•.J'=12 



I o8 KEY TO ELEMENTAR V HYDR OSTA TICS. 



18. Let w be the apparent weight of each in water. 

« 

(Hydrostatics, Art. 65, Case I.) 



rru 2-6 22 
^^^^ T = 223^' 

,7-8 n 

and-r-= ; 

1 n-w 



99 

2*6 ' 

an(i=-s=»-M; 
7*8 

22 n _ 

/. 66-n=7-8 (22 -n) ; 
.*. 6*871= 105*6, and .*. n=15^. 



19. Let w be the apparent weight of the substances in the fluids, 
s the specific gravity of the fluid. 

Then jp, 



.•.-g-=l-^, 



and 



38 

2-25 



= 3-1(7 



/. -I — 4- = 2, or,4«-«=6, and /. « = 2. 
•75 3 



k 



N 



vv 



20. Let N be the position of the fulcrum, v the volume of each 

body. 

C Then tension of string AB—v (2*7 - 1), 

and tension of string CD=v (6*1 - 1) ; 

.*. if CN=x inches, 

i; (2-7-1 )^ X ^ 

V (6-1-1) 71 -a' 

17 X 

, or, 7l-a=3«; 



6 

B 



6 

D 



Fio. 196. 51 71 -a 

/. a!=-7- inches =17} inches. 
4 



KE Y TO ELEMENTAR Y HYDROSTATICS, 109 



Examples — IX. (p. 94.) 

1. Let X and y be the lengths of the aims at the ends of which the 
gold and silver hang in each of the three cases : then 

(l)a;:y=10-4:19-4 
= 104:194 
= 52:97. 

(2) x: 2/= 10-4-1 :19'4-1 

= 9-4: 18-4 
=47 : 92. 

(3) In this case the silver is forced up by the mercury, 

and acts on the lever vertically upwards, hence 
the fulcrum must be at one end, and the gold 
between the silver and the fulcrum, and 
X : t/= 13-5 - 10-4 : 19*4 - 13*5 

= 3-1 :5-9 

= 31:59. 

2. Let X be the weight of the body in the fluid. 

r^ 1 -X 3 
Then-V^=-^; 

7 7 ' 

.•. 05=4 lbs. 
Then, by 8tcU.ic8^ Art. 102, 

height : base =3 :4. 

3. Let V be the volume of hydrogen in the balloon in cubic feet. 
Then v x '07 x 1*3= weight of hydrogen in ounces, 

and V X 1*3= weight of displaced air in ounces ; 
.*. V X 1-3=1; X -07 X 1*3 + 10 X 112 X 16 ; 
.'. V X 1-3 X -93=10 X 112 X 16 ; 

4. Let pi denote the pressure internally before the ascent, p^ the 
pressure internally after the ascent. 

Then ^2 :i>i=3 :4. 
Hence if Vi and Va be the volumes of the gas before and after the 
ascent. 



1 10 KEY TO ELEMENTAR Y HYDROSTA TICS. 

Hence one-third of the gas must have been expelled to mftintj^in 
equilibrium between the external and internal pressures. 

Also, since the pressure of the external air is only three-fourths of 
the pressure of the external air before the ascent, one-fourth of the 
whole weight must have been thrown out. 

5. The gas has been expelled to preserve equilibrium of internal 
and external pressures, and the ballast to preserve equilibrium of 
vertical pressures on the balloon. 

6. Volume of cylindrical vessel =7r . 4^ . 9 cubic inches. 
Volume of interior of cylinder =7r . 3^ . 8 cubic inches ; 

.'. volume of wood =7r (16 x 9 - 9 x 8) cubic inches = 72 tt cubic inches, 

also, volume of cylindrical mass of fluid displaced = tt . 4^ . 3 cubic in. ; 

.'. if s be the specific gravity of the wood, 

727r.s=487r; 

_48_^ 

•'•*"72"3* 

Again, let x oe the depth to which it sinks. 

Then volume of fluid displaced =7r.42 + 7r (42-3^ {x-\) 

= 77ra; + 97r; 
. *. 77rx + 97r = 727r . s = 487r ; 
.'. 7a5=39, or, a;=5f inches. 

7. No change will occur till the stone falls from the ice ; but then 
the ice will displace less water than before, and the surface of the 
water will sink. 

8. Let AB be the surface of the water outside, CD the level to 
which the water rises inside. 

p^ g Then, by Boyle's law, 

~ pressure of air in ^Z) : atmospheric pressure = jB^ : BB, 
Hence if h represent the atmospheric pressure, and x 



D be the length of BD, and y the length of DN^ 

pressure of air in AB x + y ^ 
atmospheric pressure x ' 

M N ...?+^=^or,x2 + x^=x^ + %,and.-.a:2=^y^ 

Fia, 196 h X 



KE Y TO ELEMENTAR Y HYDROSTATICS. 1 1 1 

9. The pressure is equal to the weight of (80.x 6) cubic feet of water. 
Hence if we take the weight of a cubic foot of sea water as 1000 

ounces (which is rather less than the actual weight), 

pressures (80 x 6 x 1000) ounces = 30000 lbs. 
If the box were not water-tight the water would enter it, and the 
pressure will be the same inside as outside. 

10. Let X be the number of feet to which the bag is sunk. 

The pressure on the bag will then be equal to a pressure indicated 
by a height of (a: + 34) feet on the water barometer. 

•''19 a;+34' 
.'. ic= (18 X 34) feet = 102 fathoms. 

11. Let a; be the weight of the vessel in ounces. 

Then, since the weight of a cubic foot of water is 1000 ounces, 

a; + 2- X 1000 =1000; 

8 

1000 ,-,- 
.'. a5= - = 125 ounces, 
o 

12. The surface of the fluid will rise higher than the surface of the 
water, because the specific gravity of the fluid is less than the specific 
gravity of the water. 

13. Let % be the depth in inches to which the cylinder sinks. 
Then 15 x I'l + 15 x -25=0; x 1 ; 

.-. a;=15 X 1-35 = 20*25 inches. 

14. The larger piece of cork will rise to the surface, and the length 
of the string between it and the pulley will be 2 feet, while the 
smaller piece of cork, at the end of 1 foot of string from the pulley, 
will be entirely submerged. 

Then if w and Zw be the weights of the pieces of cork, the smaller 
piece displaces water to the weight of 4t(;. 

Hence we must have an upward force =:3u; at end of longer string. 

.'. the larger piece of cork must displace water to the weight of 6w, 
end .'. it must be half immersed. 



\ 



112 KEY TO ELEMENT A R Y HYDR OSTA TICS. 



15. As the air is drawn away from the tube the water will rise in 
both branches, and if the height of the top of the tube above the 
surface of the water in the reservoirs is not too great, there will 
ultimately be a continuous column of water in both parts of the 
siphon, and then a regular flow will commence 



I 



16. Let V be the volume of each body, s^ and «2 the specific gravities 
of the bodies, p^ and p2 the specific gravities of the liquids. 

Then, «i-pi=«2~P2> o^> *i~*2=Pi~"P2> (1) 

And«i-^ = S2-Pi, or, «i-S2 = ^^-2-^ • ...... (2) 

Hence pi-p2 = ^^^ ^\ 

and thus pi = -^, and 8^ - gg = - ^* 

Let X be the part of the heavier body immersed in the final experi- 
ment. 



Then vs^^^vs^-oyvl^ — ^j ; 



X 



.-. Si-S2=--o--0'i + p2); 



. _P2 ^ . 7p2 

•• 4" 2 4' 



and thus « = -=-• 



17. ^Z> is the level of the mercury, B is the height of the water. 
Then the question to resolve is whether the weight of a column 

of mercury in height CD in the longer 
arm is greater (or not) than the weight 
of a column of water, height CRy on the 
same base, since the pressure upwards 
'^at the bottom of this tube = pressure 
- D upwards at ^ + weight of such a column 
C of water ; 
Fio, 197. .*. mercury will flow through the tube 

if CD X density of mercury be greater than CR x density of water, 

or, if zp be greater than /p', 

or, —^ be greater than — • 




KE V TO ELEMENTAR Y HYDR OSTA TICS. 1 1 3 



18. Let a; = depth of air at the top of the vessel when the water 
stands at the top of the pipe. 

The amount of water forced in at each 
stroke \b Al. 

Then, if tt be the atmospheric pressure, 
and p the atmospheric density, 

pressure of confined air 



= — 7r — (Ji- {c — x) )p + 7r, and 7r = ph ; 

X 

.*. — = 2/i-c + x, 

X 



and .'. x= 



2h±\f4h^+"c^ 



X 




l« 



Pio. 198. 



2 

Again, considering the quantity of water 
in the air-vessel and pipe, after n strokes 
of the piston. 

Aim— {c-x) A+ a(h -c + x) ; 

2AI 

19. Let x be the weight of the cylinder in lbs. 
Thenaj + 6:x=8 :5; 

.*. 5x + 30 = 8x, or, a; =10 lbs. 

20. Let «! and 53 be the specific gravities of the metals, v and w the 
measures of the volumes and weights mixed. 

Then vsj^ + vs^ = 2v . 9, 

'2w 

8f ^ 



, It? w 
and — H — 



8, 



^1 "t" ^2 — lOy. 

1 + 1 = 1 

«i «2 40' 
and hence we find s^ = 10 or 8, aiid «« = 8 or 10. 

21. Let X be the height of the cylinder. 

Then, since the air which stood at height x inches under the 
atmospheric pressure is condensed to the height of 1 inch, 

1 :a;=15 : 165 ; 

. •. .T»= -TT = 11 inches. 
lf> 

H 



i 



114 KEY TO ELEMENTAL Y HYDRO STA TICS. 



2 2 

22. The density of the air in the receiver of the air»pump is —- of -5-, 

3 3 

or -^ of what it was at first. 

•7 

The density of the air in the receiver of the condenser is after the 

3 

first stroke — of what it was at first. 

2 
The air in the barrel at the commencement of the second ascent of 

the piston is — of the density of atmospheric pressure. 

When the piston has reached a point — of the length of the barrel 

from the valve of the condenser, the air in the barrel will be of tho 
same density as the air in the receiver of the condenser. 
The valve will then open, and the air will enter. 

The air which occupied a space ( 1 + -3^ ) o^ t^e receiver will now 

occupy a space equal to the receiver ; 

.*. density will be — of the density of air in receiver ; 

«7 

11 3 33 

.•. density will be — of — , or — of original density ; 

.*. pressure in receiver of condenser : pressure in receiver of air-pump 

=?? . j4 

18* 9 

:-33:a 



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whole narrative into accord with the present state of knowledge, and also to present iStat 
facts of Roman History in a form likely to interest the Students for whom the work is 
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assist the student in points 7vhtch present 
difficulties, are ail excellent. The pas- 
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GREEK 

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A Primer of Greek Grammar, with a Preface i^ john 

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INDEX 



PAGE 

Abbott (E.), Arnold's Greek Prose 23 

— Elements of Greek Accidence . 23 

— Hellenica . . . . . 27 

— Selections from Lucian . . 24 

— and Mansfield (E.D.), Primer of 

Greek Grammar ... ax 

Acland (A.), Political Hist, of Eng. 10 

Adams (W.), Allegories ... 36 

Alford n^ean) Greek Testament . 29 

Aristophanes 35, 28 

Aristotle's Ethics .... 38 

— Organon. By J. R. Magrath . 28 
Arnold ^T. K.) Cornelius Nepos . ao 

— Crusius' Homeric Lexicon . . 26 

— Demosthenes .... 27 
-— Eclogae Ovidianae ... 18 

— Eng. Greek Lexicon . . 29 

— First French Book and Key . 34 

— First German Book and Key . 33 

— First Greek Book and Key . 22 
; revised by F. D. Morice . 22 

— F'irst Hebrew \io6k. and Key . 35 

— First Verse Book and Key . 16 

— (Jreek Accidence ... 22 

— (ireek Prose Comp. and Key . 92 
revised by E. Abbott . . 23 

— Henry's First I^tin and Key . 16 
rexised by C. G. Gepp . . 16 

— Homer's Iliad .... 26 

— I^tin Prose Comp. and Key . 17 
revised by G. G. Bradley . 17 

— Mad Nag's Greek Syntax . . 22 

— Sophocles '28 

Bakkk (W.), Manual of Devotion . 
Harbier (P.), French Reader . 
Barrett (W. A.), Chorister's Guide . 

— Form and Instrumentation 
Belcher (H.), Livy, Book II. . 
Bennett (G. L.X Caisar's Gallic War 

— Easy Latin Stories and Key 

— First Latin Exercises . 

— First Latin Writer and Key 

— Latin Accidence .... 

— Second Latin Writer and Key . 

— Vergil, Selections from 

— Works by 

Bigg (C), I^tin Prose Exercises . 

— Thucydides, Books 1. 11. . 
Blunt (J. H.), Household Theology 

— Ke>-s to Christian Knowledge . 
}k>wen (E. E.), Campaigns of Napoleon 
Bradl«y(G. G.), AmolcTs Latin Prose 
Bridge (C.XFrcndi Literature 

Bright CJ. F.). History of England . 
Buuding Construction. Notes on . 
Ikirton (J.), English Grammar 

CiSSAR . . . . . 15, z6, x8 
Calvert (E.), Selectioiis from Livy . ao 
Can- (A.), Notes on St. Luke . . ag 



3a 

34 
35 
35 
19 
15 
15 
14 
14 
14 
14 
17 
38 
18 
28 
3a 
31 
34 
17 



Catena Gassiccn^m 
Cicero de Amicitia . . ' . 
Clarke (A. D.), Examination Papers 
Cornelius Nepos. By T. K. Aniold 
Cornish (F. Wjl, Oliver Cromwell . 
Crake (A. D.), History of the Church 

— Edwy the Fair . , . . 

— Alfgar the Dane 

— Ki\'al Heirs . . . . 
Creighton {LX First Hist, of Eng. 
-«- Historical Biographies 

— Stories from Eng. Hist. 
Crusius (G. CX Homeric Lexicon . 
Curteis (A. M.), The Roman Empire 

Dallin (T.X Materials and Models 
Davys (Bishop), History of England 
Dawe (C. J. S.), Latin Exercise Bk 

— Study of Church Catediism 
Demosthenes .... 



II 
5 



English School-Classics 
Euripides, Scenes from . 

FiKF.i) (W.), Stones of the Temple 
Firman (F. B.JL Zeugma . 
Fradersdorif, Eng.-Greek Lex. 

Gantillon(P.J.F.), Exam. Papers 
Gedge (I. W.), Com. to Braver Book 
Gepp (C. G.), AmoW's Henry's First 
I^tin Book .... 

— I^tin ElegUc Verse . 
r- I..atin-English Dictionary . 

— VirgU 

Girdlestone (W. H.), Arithmetic . 
Goethe's Faust .... 
Goolden(W. T.), Intra to Chemistry 
Goulbum (Dean), Confinnati<Mi , 
Green (A. H .), Geology for Student! 

— (W. C^ Aristophanes 
Grenfell (E. FX German Exercises 
Gross (E. J.), Algebra, Part 11. ' . 

— Kinematics and Kinetics . 

Hardy (E. G.). Antiq. of Greece . ay 
Harrison (J. E.) Myths of the Odyssey 27 

HaufTs Stories, Selecticms from . 33 

Heatley (H. R.), (^datim . • ■ X5 

Hellenica, Essays . . . . 37 

Herodotus, Stones frtMn, Phillpotts . 34 

— By H. G. Woods ... 37 
Hertz (H. A.), Short Readings . 5 
Heslop (G. H.), Demosthenes . 27 
Hcygate (W. E.), Allegories . . 36 
Historical Biographies ... 10 
Historical Handbooks ... 9 
Holmes (A0» Demosthenes . . aj 

— Rules of^ Latin Proncuiciation . 15 

Homer's Iliad a6 

Horace. By J. M. Marshall . ao 

loPHON a4 

Isocrates. By J. E. Sandys . 26 



PAGB 

!S 

»3 
ao 

zo 

so 

36 
36 
36 
8 
xo 

3« 
a6 



»9»a4 
xo 

X5 

32 

37 

6,7 

85 

36 

•4 
*9 

»9ia4 
3x 

x6 
x6 

19 
17 
13 
33 
xz 

3x 

XX 

a8 

33 
xa 

xa 



s • * - 



INDEX, 



\ 



PAGB 

BBB (R. C.)} Sophocles ... 28 
ennings (A. C), Ecclesia Anglicana 32 
uveiuu. By G. A. Simcox . . ao 



Kkys to Chbistian Knowlbdgb . 

— List of 

Kitchener (F. A.), A Year's Botany 
Kingdon (H. N.), Gradatim . 

Latin Tbxt Books 

La Fontaine's Pabl)£k By P. Smith 



Lang (L. B),Geography for Beginners 

)n(j.),/ 
Laun (Van, H), French Selections 



, At Home and Abroad 



Laughtor 

Laun (Vaw, **/, a^w»^,u wwt«^^...v 
E^sin^'s Fables. By F. Storr 
Libraries, Books suitable for . 
Livyi Selections from 
Lucian, Selections from . 

Macbbth (J.), Church Principles 
Madvig's Greek. Syntax . 
MagrathCJ. R.), Aristotle's Organon 
Mann (J. SO* Aitiquities of Greece 
Mansfield (£. D.), Latin Sentence . 

— Primer of Greek Syntax .^ 
Manuals of Religious Instruction . 
Marshall (J. M.), Horati Opora 
Merryweather (J. H.), Caesar . 
Moberly(C.£.), Alexander the Great 

— Geography 

— Shakspere's Play; . . 

— Xen(^hon's Memorabilia . 
Moore (£.), i^ristotle's Ethics . 
Moore cE. H.), Greek Method 
Morice(F. D.), Stories in Attic Greek 
Morshead (E.D.), Goethe's Faust . 
Mullins (W. E.), HauflTs Stories . 

Napoleon's Campaigns 
Norris CF. P.), New Testament 

— Confirmation . 

— Keys to Christian Knowledge . 
-^ Rudiments of Theology 

OviDiANiS Eclogae. By Arnold 
Ovid, Stories from. By R. W. Taylor 

Papillon (T. L.), Terenti Comoediae 
Pearson TC. H^, English History 
Percival (J.), Helps for School Life 
Persius. By A. Pretor . 
Phillpo'tts (J. S.). Homer's Iliad 

— Shakspere's 1 empest 

— Stories from Herodotus 
PoweU (F. York)j English History 
Prefepr (A.), Persii Satirae 
Priestland (E.), Greek Prepositions 
Prizes, Books suitable for 

RANSOMK(C0f Political Hist of Eng 

— History of^tlfi Romans 
Reynolds (S. H.% IBad of Homer 
Richardson (G.), Conic Sections 
Rigg (A.), Intro, to Chemistry 
Ritchie (F.), First Steps in Latin 

— Practiced Greek Method . 
Rivington's Mathematical Series 
Rogers (J. E. T.), Aristotle's Ethics 



12, 



31 

4 

XI 

XS 

16 
34 
35 
35 
34 
33 
36 
20 

24 

32 
22 

28 

27 

19 
21 

31 
20 
18 
25 
35 
5 
25 
28 
21 

25 
33 
33 

34 
32 
31 
31 
32 

18 
18 

20 
9 
31 
20 
26 

5 
24 

8 

20 
23 

10 

9 
26 

12 
II 

15 
21 
13 
28 



PAGB 

Sanderson (L.), Zeugnoa . . 24 

Sandys Q. £.), Isocratis Orationes 26 
Sargent Q.), Materials and Models . 19, 34 

Schoemann s Antiquities of Greece 37 

Shakspere's Plays . ... 5 

Sidgwick (A.), Cicero de Amicitia . x8 

— First Greek Writer . v • • *« 

— Greek Prose Compositidb . . m 

— Homer's Iliad . . . . a6 

— Scenes from Greek Plays . .as 

— Works by . . . . . 38 
Simcox (G. A.), Juvenalis Satirae . so 

— Thucydides . •. •. • "S 

— (W. H.), Taciti Historise , ao 
Smith (J. Hamblin), The Acts . ag 

— Algebra and Key . . . xa 
-^ Algebra, Exercises on . . za 

— Arithmetic and Key . . . xa 

— Book of .jEnnnci^tioiis . . 13 

— English Gramniar ... . 5 

— Geometry and Key . . . .^.13 

— Greek Graminar . . "t* . "."J13 

— Heat, The Study of . . . X3 

— Hydrostatics dmd Key . . xa 

— Latin Gramnjar . . . . z6 
Prose Composition and Key . x8 

— Statics and Key . . ... xa 

— St. Mark's Gospel ... 29 

— Trigonometry and Key . . la 

— Works by 37 

— f P. Bowden), La Fontaine's Fables 34 

— (P. v.), English Institutions . 9 

— iR. Prowde), Latin Prose Ex. . 15 
Sophocles ..... b8 
Storr (F.), iEneid of Vergil . . 17 

— Greek Veri)s^ 33 

— HauflTs Sfones. •'.-''•, . . 33 
I — Lessing's FalJte$'' • • • 33 

Tacitus. By W. H. Simcox . . ao 

I Tancock (C. C), Caesar . . x8 
, Taylor (R. W.), Short Greek Syntax 
i — Stones from Ovid 

— Xenophon's Agesilaus 

— Xenophon's Anabasis 

— Works by . •. . 
, Terence. By T. L. Papillon . 
. Tidmarsh (W.) English Grammar . 

Thucydides .*.... 
; Turner (E. J.) Goethe's Faust 



as 

as 
38 
ao 

5 

aS 

33 



I Vecquerav (J.), German Accidence 33 
Vergil t6, X7 



Waite (R.), Duke of Wellington 
Way of Life .... 
Whitelaw, Shakspere's Coriolanus 
Willert (P. F.), Reign of Lewis XI 
Wilson's Modem English Law 
Woods (H. G.), Herodoti Historia 
Wordsworth (Bp.), Greek Testsunent 
Wormell, Principles of I)ynamics 
Worthington's Practical Physics 



•*. 



Xbnophon 



xo 
3« 

5 

9 

9 

•7 

•9 

X3 

XX 

•5 



RIVINGTONS' EDUCATIONAL LIST 



A rnolcPs L atin Prose 

Compontum, By G. 6. Bradley. 

5*. 

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[The original Edition is still on sale.] 

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By G. L. Bbnnbtt. ^. 6d. 

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Beginners. By G. L. Bennett. 

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By G. In Bennett. 2J. 

Selections from Vergil, 

By G. L. Bennett, is. Gd. 

Virgil Georgics, Book iv. 

By C. G. Gepp. i*. (xL 

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WICK. 

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Stories from Ovid in 

Elegiac Verse. By R. W. Taylor. 
3*. td. 

Stories from Ovid in 

Hexameter Verse. By R. W. Tay- 
lor. 2f . td. 



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